/*! 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 110 Let \(t=\) the amount of sales t... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(t=\) the amount of sales tax a retailer owes the government for a certain period. The article "Statistical Sampling in Tax Audits" (Statistics and the Law, 2008: \(320-343\) ) proposes modeling the uncertainty in \(t\) by regarding it as a normally distributed random variable with mean value \(\mu\) and standard deviation \(\sigma\) (in the article, these two parameters are estimated from the results of a tax audit involving \(n\) sampled transactions). If \(a\) represents the amount the retailer is assessed, then an under- assessment results if \(t>a\) and an over-assessment results if \(a>t\). The proposed penalty (i.e., loss) function for over- or under-assessment is \(\mathrm{L}(a, t)=t-a\) if \(t>a\) and \(=k(a-t)\) if \(t \leq a(k>1\) is suggested to incorporate the idea that over-assessment is more serious than under- assessment). a. Show that \(a^{*}=\mu+\sigma \Phi^{-1}(1 /(k+1))\) is the value of \(a\) that minimizes the expected loss, where \(\Phi^{-1}\) is the inverse function of the standard normal cdf. b. If \(k=2\) (suggested in the article), \(\mu=\$ 100,000\), and \(\sigma=\$ 10,000\), what is the optimal value of \(a\), and what is the resulting probability of over-assessment?

Short Answer

Expert verified
The optimal value of assessment \(a^*\) is $95,700 and the probability of over-assessment is approximately 0.333.

Step by step solution

01

Understanding the Loss Function

The loss function is defined as \( L(a, t) = t - a \) if \( t > a \) and \( L(a, t) = k(a-t) \) if \( t \leq a \), where \( k > 1 \) to account for over-assessment being more serious.
02

Calculating Expected Loss

The expected loss \( E[L(a, t)] \) involves integrating the loss function over the probability density function of \( t \), considering separately the cases \( t > a \) and \( t \leq a \).
03

Expressing Probabilities with the Standard Normal Distribution

Since \( t \) is normally distributed with mean \( \mu \) and standard deviation \( \sigma \), we can standardize it by using \( Z = \frac{t-\mu}{\sigma} \), where \( Z \) is a standard normal random variable.
04

Minimizing the Expected Loss

The optimal \( a^* \) must satisfy \( E[L(a, t)] \) minimized. Through optimization, we find \( a^* = \mu + \sigma \Phi^{-1}\left(\frac{1}{k+1}\right) \).
05

Applying k = 2 to Find Optimal a

Plugging in \( k = 2 \), \( \mu = 100,000 \), and \( \sigma = 10,000 \), the optimal \( a \) is calculated by \( a^* = 100,000 + 10,000 \Phi^{-1}\left(\frac{1}{3}\right) \).
06

Calculating \( \Phi^{-1}(\frac{1}{3}) \)

Using a standard normal distribution table or calculator, \( \Phi^{-1}(\frac{1}{3}) \approx -0.43 \).
07

Calculate Optimal a Value

Substitute \( \Phi^{-1}(\frac{1}{3}) \approx -0.43 \) into the formula: \( a^* = 100,000 + 10,000(-0.43) = 95,700 \).
08

Calculating Probability of Over-Assessment

The probability of over-assessment corresponds to the probability that \( t \leq a^* \), which is \( \Phi\left( \frac{a^* - \mu}{\sigma} \right) = \Phi(-0.43) \approx 0.333 \).

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics. It describes how values of a variable are distributed. This distribution is symmetric around its mean and characterized by its bell-shaped curve.
Most values cluster around a central peak, and probabilities of values further away from the mean taper off equally in both directions. The normal distribution is defined by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).
- The mean is the average value of the data set.- The standard deviation measures how much variation or dispersion exists from the mean.
In the problem about tax audits, the amount of sales tax a retailer owes, \(t\), is assumed to be normally distributed. This assumption allows us to apply statistical techniques to model and predict tax assessments and potential discrepancies.
Expected Loss
Expected loss quantifies the average loss we anticipate under uncertainty, considering all possible outcomes and their probabilities.
In our exercise, the loss function - is \(L(a, t) = t - a\) if there's an under-assessment (actual tax \(t\) exceeds assessed tax \(a\)) - and \(L(a, t) = k(a-t)\) if there's an over-assessment, with \(k > 1\) emphasizing the severity of over-assessment.
To compute expected loss, these potential losses are averaged over all scenarios, weighted by their probabilities. Integrating this function over the probability distribution of \(t\) gives us the expected value, essentially predicting what loss we can expect given the normal distribution of \(t\). This helps determine the decision that minimizes risks for stakeholders.
Optimization in Statistics
Optimization in statistics involves finding parameters or values that minimize or maximize a statistical function. In the context of tax audits, our goal is to find the optimal assessment value \(a^*\) that minimizes the expected loss function.
- This involves calculus, specifically setting derivative(s) of the loss function's expected value to zero to find critical points, indicating potential minimum or maximum values.
- By examining these points, it’s possible to determine which one minimizes the expected loss.
The solution shows that the optimal \(a^*\) is computed using the formula \(a^* = \mu + \sigma \Phi^{-1}(\frac{1}{k+1})\). This reflects in simple terms, a balance between under-assessment and over-assessment as dictated by the penalty factor \(k\), offering the best possible assessed tax to minimize potential discrepancies.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution, with a mean of zero and a standard deviation of one. Any normal distribution can be transformed into this standard form through a process known as standardization.
- The transformation involves subtracting the mean \(\mu\) from the variable and dividing by the standard deviation \(\sigma\). This yields a new variable \(Z = \frac{t-\mu}{\sigma}\).
- The standard normal distribution allows for the use of standard tools like Z-tables to find probabilities and quantiles.
In this problem, the value \(\Phi^{-1}(\frac{1}{k+1})\) is found using the standard normal distribution's inverse cumulative distribution function (CDF).
The solution uses this to determine the specific point at which to set the assessment to minimize expected loss effectively, exemplifying its practicality in statistical sampling and decision-making.

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