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Please read the Focus Problem at the beginning of this chapter. Recall that Benford's Law claims that numbers chosen from very large data files tend to have " \(1 "\) as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with " 1 " as the leading digit is about \(0.301\) (see the reference in this chapter's Focus Problem). Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of \(n=215\) numerical entries from the file and \(r=46\) of the entries had a first nonzero digit of \(1 .\) Let \(p\) represent the population proportion of all numbers in the corporate file that have a first nonzero digit of \(1 .\) j. Test the claim that \(p\) is less than \(0.301\). Use \(\alpha=0.01\). ii. If \(p\) is in fact less than \(0.301\), would it make you suspect that there are not enough numbers in the data file with leading 1 's? Could this indicate that the books have been "cooked" by "pumping up" or inflating the numbers? Comment from the viewpoint of a stockholder. Comment from the perspective of the Federal Bureau of Investigation as it looks for money laundering in the form of false profits. iii. Comment on the following statement: "If we reject the null hypothesis at level of significance \(\alpha\), we have not proved \(H_{0}\) to be false. We can say that the probability is \(\alpha\) that we made a mistake in rejecting \(H_{0} . "\) Based on the outcome of the test, would you recommend further investigation before accusing the company of fraud?

Short Answer

Expert verified
Rejecting \( H_0 \) suggests \( p < 0.301 \), potentially indicating data manipulation. Recommend further investigation.

Step by step solution

01

Define Hypotheses

Let the null hypothesis be \( H_0: p = 0.301 \) and the alternative hypothesis be \( H_1: p < 0.301 \). We want to test if the proportion of numbers with the first digit as \( 1 \) is less than \( 0.301 \).
02

Calculate Sample Proportion

The sample proportion is calculated as follows: \( \hat{p} = \frac{r}{n} = \frac{46}{215} \approx 0.2135 \).
03

Find the Test Statistic

The test statistic for a proportion can be calculated using the formula: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 = 0.301 \). Substituting the values, we get \( z = \frac{0.2135 - 0.301}{\sqrt{\frac{0.301(1-0.301)}{215}}} \approx -2.716 \).
04

Determine Critical Value and Compare

For a one-tailed test with \( \alpha = 0.01 \), the critical value from the standard normal distribution table is approximately \( -2.33 \). Since \( z = -2.716 \) is less than \( -2.33 \), we reject the null hypothesis \( H_0 \).
05

Interpretation of Result

Rejecting \( H_0 \) suggests that there is significant evidence that \( p \) is less than \( 0.301 \). This discrepancy from Benford's Law may raise suspicions that the data has been manipulated.
06

Investigate Implications

As a stockholder, noticing that there are fewer leading 1's might raise concerns about the authenticity of the data, suggesting that numbers might be inflated. From a law enforcement perspective, such deviations could warrant a deeper investigation into potential financial misstatements.
07

Consideration for Further Investigation

The statement emphasizes that rejecting \( H_0 \) does not prove it false, only that there is significant evidence against it. A 1% chance remains that the test result is due to random sampling variability. Hence, it is prudent to recommend further investigation before concluding any fraudulent activities.

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Key Concepts

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

Benford's Law
Benford's Law is a fascinating statistical principle often used in data analysis and forensic accounting. It states that in many naturally occurring data sets, the first nonzero digit is more likely to be small. In particular, the number 1 appears as the first digit about 30.1% of the time.

This might seem counterintuitive at first because we might expect each digit from 1 to 9 to appear equally likely. However, Benford's Law is often validated in real-life datasets like population numbers, physical and mathematical constants, or even financial records.
  • Usage: Benford's Law can help in detecting anomalies or fraudulent activities in data sets. If the data deviates significantly from the expected distribution, it could suggest manipulation.
  • Application: Auditors and analysts use this law to check for consistency in financial records, making it a powerful tool in forensic accounting.
Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It represents the default position or claim that there is no effect or no difference. In the context of our problem, the null hypothesis is that the population proportion, \( p \), of numbers starting with the digit "1" is 0.301.

Forming a null hypothesis is essential when you want to test whether an observed effect could be due to chance. The hypothesis you want to gather evidence against is what you're typically testing.
  • Purpose: To provide a baseline for testing the validity of an alternate hypothesis, which suggests a new effect or difference.
  • Outcome: If the null hypothesis is rejected, it provides enough evidence to support the alternate hypothesis.
Sample Proportion
A sample proportion is a statistical metric that represents the proportion of occurrences of a specific characteristic in your sample. It's an estimate of the population proportion. In our exercise, the sample proportion \( \hat{p} \) was calculated as 0.2135 since 46 out of 215 numbers in the sample had "1" as the leading digit.

Calculating the sample proportion is critical in hypothesis testing to determine how your sample represents the broader population.
  • Calculation: \( \hat{p} = \frac{r}{n} = \frac{46}{215} \approx 0.2135 \), where \( r \) is the number of successful occurrences and \( n \) is the total sample size.
  • Usefulness: The sample proportion is used to compute the test statistic, which is then compared against the null hypothesis to draw conclusions.
Test Statistic
The test statistic is a standardized value used in hypothesis testing to decide whether to reject the null hypothesis. It tells us how far our sample statistic (sample proportion in this case) is from the hypothesized population proportion, measured in units of standard error.

In simple terms, the test statistic helps determine if the observed sample proportion provides enough evidence against the null hypothesis.
  • Calculation: The formula is \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 \) is the hypothesized population proportion, \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.
  • Decision-Making: The computed test statistic \( z \) value is then compared against the critical value from the standard normal distribution to determine whether to reject the null hypothesis.
  • In Our Case: A test statistic of approximately -2.716 indicated rejection of the null hypothesis since it was less than the critical value of -2.33 at a 0.01 significance level.

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Most popular questions from this chapter

A random sample of \(n_{1}=16\) communities in western Kansas gave the following information for people under 25 years of age. \(x_{1}:\) Rate of hay fever per 1000 population for people under 25 \(\begin{array}{rr}98 & 90 \\ 125 & 95\end{array}\) 120 125 \(\begin{array}{ll}0 & 128 \\ 25 & 117\end{array}\) \(\begin{array}{ll}28 & 92 \\\ 17 & 97\end{array}\) 123 \(\begin{array}{ll}112 & 93 \\ 127 & 88\end{array}\) \(\begin{array}{lllll}5 & 125 & 117 & 97 & 12\end{array}\) A random sample of \(n_{2}=14\) regions in western Kansas gave the following information for people over 50 years old. \(x_{2}:\) Rate of hay fever per 1000 population for people over 50 \(\begin{array}{ll}95 & 110 \\ 79 & 115\end{array}\) 10 1 \(\begin{array}{llllr}101 & 97 & 112 & 88 & 110 \\ 100 & 89 & 114 & 85 & 96\end{array}\) (Reference: National Center for Health Statistics.) i. Use a calculator to verify that \(\bar{x}_{1}=109.50, s_{1}=15.41, \bar{x}_{2}=99.36\), and \(s_{2} \approx 11.57\) ii. Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\alpha=0.05\).

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Let \(x\) be a random variable that represents the \(\mathrm{pH}\) of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean \(\mathrm{pH}\) level of the blood.

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