/*! 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 11 Benford's law Faked numbers in t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{3}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ A forensic accountant who is familiar with Benford's law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data. $$ \begin{array}{lcrrrrrrrr} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Count: } & 61 & 50 & 43 & 34 & 25 & 16 & 7 & 8 & 6 \\ \hline \end{array} $$ (a) Are these data inconsistent with Benford's law? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

Short Answer

Expert verified
The data is inconsistent with Benford's law if the computed chi-square statistic exceeds the critical value. A Type I error falsely accuses fraud; Type II misses fraud. Type I error is potentially more serious.

Step by step solution

01

State Hypotheses

To test if the data follows Benford's law, we state the null hypothesis, which is that the observed distribution of first digits follows Benford's law. Formally, the null hypothesis is \( H_0: P(X = i) = p_i \) for all \( i = 1, 2, \ldots, 9 \), where \( p_i \) are probabilities given by Benford's law. The alternative hypothesis \( H_a \) is that the distribution does not follow Benford's law: \( H_a: P(X = i) eq p_i \) for some \( i \).
02

Calculate Expected Counts

Calculate the expected counts under the null hypothesis using the formula \( E_i = n \cdot p_i \), where \( n \) is the total number of observations (250) and \( p_i \) are the probabilities according to Benford's law. Perform this calculation for each digit.
03

Compute Chi-square Statistic

Use the chi-square statistic formula: \[ \chi^2 = \sum_{i=1}^9 \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed count for digit \( i \) and \( E_i \) is the expected count. Compute this value using your observed and expected counts.
04

Determine Degrees of Freedom

The degrees of freedom for a chi-square goodness-of-fit test is the number of categories minus 1. Here, there are 9 first-digit categories, so the degrees of freedom is \( 9-1=8 \).
05

Find Critical Value and Compare

Using a chi-square distribution table or calculator, find the critical value for \( \alpha = 0.05 \) and 8 degrees of freedom. Compare the computed chi-square statistic to the critical value to decide whether to reject the null hypothesis.
06

Interpret Results

If the chi-square statistic is greater than the critical value, reject the null hypothesis, indicating the data is inconsistent with Benford's law. Otherwise, do not reject the null hypothesis, indicating no evidence of inconsistency.
07

Describe Type I and Type II Errors

A Type I error occurs if we conclude the data does not follow Benford's law when it actually does, possibly leading to a false accusation of fraud. A Type II error happens if we conclude the data follows Benford's law when it does not, potentially missing fraudulent activity. In this setting, a Type I error might be more serious if it falsely accuses a company of fraud.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chi-square Test
The Chi-square test is a powerful statistical tool often used to compare observed data with data we would expect to obtain according to a specific hypothesis.
When the exercise asks whether data aligns with Benford's Laws, it requires the Chi-square test.
It checks if observed frequencies differ significantly from what Benford's Law predicts.

Here's how it works in simple terms:
  • First, we calculate the expected counts for each digit using Benford's Law probabilities.
  • Next, compare the observed counts (from the sample data) with the expected counts using the Chi-square formula: \[ \chi^2 = \sum_{i=1}^9 \frac{(O_i - E_i)^2}{E_i} \]where \( O_i \) is the observed count, and \( E_i \) is the expected count for each digit \( i \).
  • The Chi-square value obtained tells us the extent to which observed counts deviate from expected counts.
This value is then compared to a critical value from the Chi-square distribution (dependent on the number of categories and significance level \( \alpha \)).
If our calculated Chi-square is higher than this critical value, our null hypothesis (that data follows Benford's Law) is likely incorrect.
Type I and Type II Errors
Type I and Type II errors are vital concepts in hypothesis testing. Understanding these errors helps us gauge the reliability of our conclusions from statistical tests.
- **Type I Error**: Occurs when we wrongly reject a true null hypothesis. In our context, this means concluding that the data does not follow Benford's Law when it actually does.
This can lead to unjustly suspecting or accusing a company of fraud.

- **Type II Error**: Happens when we fail to reject a false null hypothesis. Here, it means deciding that the data follows Benford's Law when, in fact, it does not. This error could let fraudulent activities go undetected.

To minimize these errors, statisticians set an acceptable level of potential risk called the significance level (\( \alpha \)). This balances the chances of making either error. However, real-life situations often present trade-offs; deciding which error is more impactful depends on the specific circumstances.
Forensic Accounting
Forensic accounting is the practice of using accounting, auditing, and investigative skills to examine financial statements and activities thoroughly.
This field plays a crucial role in detecting and preventing fraud.

Using tools like Benford's Law and the Chi-square test, forensic accountants can identify inconsistencies or anomalies in financial records.
These anomalies might suggest manipulations or misreporting of data.
  • Benford's Law, for example, is frequently employed to scrutinize the authenticity of financial data through its expected distribution of first digits.
  • Abnormalities in these distributions can be red flags warranting further investigation.
Forensic accounting requires not only technical accounting knowledge but also an understanding of psychological and financial norms to uncover hidden truths in financial data.
Goodness-of-fit Test
The Goodness-of-fit test is a cornerstone in statistical comparison tasks, determining how well observed samples align with expected distributions.
In the case of Benford's Law, the test checks if the collected data follows a certain expected probability distribution.

This involves:
  • Identifying the expected pattern or model, like Benford's Law.
  • Collecting observed data, such as the actual distribution of first digits in financial records.
  • Calculating the Chi-square statistic to measure the closeness of fit between observed and expected distributions.
When applying this test, statisticians decide, based on the Chi-square value and degrees of freedom, if data observed provides a good fit to Benford's predicted pattern.
If the fit is poor, it suggests that some other factors may be influencing the data's distribution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Attitudes toward recycled products Some people believe recycled products are lower in quality than other products, a fact that makes recycling less practical. Here are data on attitudes toward coffee filters made of recycled paper from a random sample of adults: \(^{21}\) $$ \begin{array}{lcc} {\text { Recycled coffee filter status }} \\ \text { Quality rating } & \text { Buyers } & \text { Nonbuyers } \\ \text { Higher } & 20 & 29 \\ \text { Same } & 7 & 25 \\ \text { Lower } & 9 & 43 \\ \hline \end{array} $$ Make a well-labeled bar graph that compares buyers' and nonbuyers' opinions about recycled filters. Describe what you see.

Going Nuts The UR Nuts Company sells Deluxe and Premium nut mixes, both of which contain only cashews, brazil nuts, almonds, and peanuts. The Premium nuts are much more expensive than the Deluxe nuts. A consumer group suspects that the two nut mixes are really the same. To find out, the group took separate random samples of 20 pounds of each nut mix and recorded the weights of each type of nut in the sample. Here are the data: \({ }^{18}\) $$ \begin{array}{lcc} {\text { Type of mix }} \\ \text { Type of nut } & \text { Premium } & \text { Deluxe } \\ \text { Cashew } & 6 \mathrm{lb} & 5 \mathrm{lb} \\ \text { Brazil nut } & 3 \mathrm{lb} & 4 \mathrm{lb} \\ \text { Almond } & 5 \mathrm{lb} & 6 \mathrm{lb} \\ \text { Peanut } & 6 \mathrm{lb} & 5 \mathrm{lb} \end{array} $$ Explain why we can't use a chi-square test to determine whether these two distributions differ significantly.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

Python eggs How is the hatching of water python eggs influenced by the temperature of the snake's nest? Researchers randomly assigned newly laid eggs to one of three water temperatures: hot, neutral, or cold. Hot duplicates the extra warmth provided by the mother python, and cold duplicates the absence of the mother. Here are the data on the number of eggs that hatched and didn't hatch: \({ }^{15}\) $$ \begin{array}{lccc} {\text {Water Temperature }} \\ \text { Hatched? } & \text { Cold } & \text { Neutral } & \text { Hot } \\ \text { Yes } & 16 & 38 & 75 \\ \text { No } & 11 & 18 & 29 \\ \hline \end{array} $$ (a) Compare the distributions of hatching status for the three treatments. (b) Are the differences between the three groups statistically significant? Give appropriate evidence to support your answer.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.