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According to Benford's law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below.Test for goodness-of-fit with the distribution described by Benford's law. $$\begin{array}{l|c|c|c|c|c|c|c|c|c} \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \begin{array}{l} \text { Benford's Law: Distribution } \\ \text { of Leading Digits } \end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\ \hline \end{array}$$ Exercise 21 lists the observed frequencies of leading digits from amounts on checks from seven suspect companies. Here are the observed frequencies of the leading digits from the amounts on the most recent checks written by the author at the time this exercise was created: \(83,58,27,21,21,21,6,4,9 .\) (Those observed frequencies correspond to the leading digits of \(1,2,3,4,5,6,7,8,\) and \(9,\) respectively.) Using a 0.01 significance level, test the claim that these leading digits are from a population of leading digits that conform to Benford's law. Does the conclusion change if the significance level is \(0.05 ?\)

Step by step solution

01

State the Hypotheses

State the null and alternative hypotheses. The null hypothesis (H鈧�): The observed frequencies follow Benford's law.The alternative hypothesis (H鈧�): The observed frequencies do not follow Benford's law.

02

Calculate the Expected Frequencies

Calculate the expected frequencies for each leading digit based on Benford's law and the total number of checks observed.Total checks = 250 (sum of observed frequencies)Expected frequencies:For leading digit 1: Expected frequency = 0.301 脳 250 = 75.25 For leading digit 2: Expected frequency = 0.176 脳 250 = 44 For leading digit 3: Expected frequency = 0.125 脳 250 = 31.25 For leading digit 4: Expected frequency = 0.097 脳 250 = 24.25 For leading digit 5: Expected frequency = 0.079 脳 250 = 19.75 For leading digit 6: Expected frequency = 0.067 脳 250 = 16.75 For leading digit 7: Expected frequency = 0.058 脳 250 = 14.5 For leading digit 8: Expected frequency = 0.051 脳 250 = 12.75 For leading digit 9: Expected frequency = 0.046 脳 250 = 11.5

03

Compute the Chi-Square Test Statistic

Compute the chi-square (蠂虏) test statistic using the formula: \[ 蠂虏 = \sum_{i=1}^{n} \frac {(O_i - E_i )虏}{E_i} \]Where 鉄∣_i鉄� are the observed frequencies and 鉄∣_i鉄� are the expected frequencies.鉄ㄏ嚶测煩 = ((83 - 75.25)虏 / 75.25) + ((58 - 44)虏 / 44) + ((27 - 31.25)虏 / 31.25) + ((21 - 24.25)虏 / 24.25) + ((21 - 19.75)虏 / 19.75) + ((21 - 16.75)虏 / 16.75) + ((6 - 14.5)虏 / 14.5) + ((4 - 12.75)虏 / 12.75) + ((9 - 11.5)虏 / 11.5)鉄ㄏ嚶测煩 = 0.779 + 4.273 + 0.578 + 0.436 + 0.084 + 1.089 + 4.976 + 5.018 + 0.543 = 17.776

04

Determine the Degrees of Freedom

Calculate the degrees of freedom (鉄╠f鉄�):鉄╠f鉄� = (number of leading digits - 1)鉄╠f鉄� = 9 - 1 = 8

05

Compare Against Chi-Square Distribution

Compare the computed chi-square value to the critical value from chi-square distribution tables for 鉄╠f鉄� = 8 at the significance levels of 0.01 and 0.05.For 鉄ㄎ扁煩 = 0.01, critical value 鉄ㄏ嚶测煩 (8) = 20.090For 鉄ㄎ扁煩 = 0.05, critical value 鉄ㄏ嚶测煩 (8) = 15.507Since 17.776 < 20.090 but 17.776 > 15.507, this leads to rejection of H鈧� at the 0.05 significance level but not at 0.01 level.

06

Conclusion

Based on the chi-square test, at the significance level of 0.01, we do not reject the null hypothesis; hence, we cannot conclude that the leading digits do not follow Benford's law. However, at the significance level of 0.05, we reject the null hypothesis; thus, suggesting the leading digits do not conform to Benford's law.

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Key Concepts

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

goodness-of-fit test

A goodness-of-fit test is a statistical test used to determine how well sample data fit a distribution from a population with a normal distribution. In this case, we want to test if the leading digits of some data follow Benford's Law. The goodness-of-fit test compares the observed data with the data we would expect if the null hypothesis is true. If the differences between the observed and expected data are too large, we can reject the null hypothesis.

chi-square test

The chi-square test, pronounced as 'kai-square,' is a statistical method used in a goodness-of-fit test to see if the observed data matches the expected data. The formula for the chi-square statistic is: \[ 蠂虏 = \sum_{i=1}^{n} \frac {(O_i - E_i )虏}{E_i} \]Where \( O_i \) is the observed frequency for each category, and \( E_i \) is the expected frequency for each category based on Benford's Law. In our example, we calculate the chi-square test statistic by summing up the differences between observed and expected frequencies squared, divided by the expected frequencies for each of the leading digits from 1 to 9.

null hypothesis

The null hypothesis, often denoted by \( H_0 \), is a statement assumed to be true for the purposes of testing. In the context of our chi-square test for Benford's Law, the null hypothesis is that the observed leading digit frequencies follow Benford's Law. The alternative hypothesis \( H_1 \) is that the frequencies do not follow Benford's Law. The goal of the test is to determine whether there is enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis.

significance level

The significance level, often denoted by \( \, 伪 \, \), is the threshold at which you decide whether an observed result is statistically significant. Common significance levels are 0.01 and 0.05. In our test, these correspond to a 1% and 5% likelihood, respectively, of rejecting the null hypothesis when it is true (Type I error). If the calculated chi-square statistic is greater than the critical value for a given significance level and degrees of freedom, we reject the null hypothesis. In our example, at the 0.05 significance level, we reject \( H_0 \), while at the 0.01 level, we do not.

leading digits distribution

The leading digits distribution according to Benford's Law states that in many naturally occurring collections of numbers, the first digits are not uniformly distributed. Instead, smaller digits occur more frequently as leading digits. For example, the number '1' appears as the leading digit about 30.1% of the time, '2' appears 17.6% of the time, and so on, decreasing to '9', which appears 4.6% of the time. This distribution forms the basis for our expected frequencies in the chi-square goodness-of-fit test.