Benford’s Law. 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. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits	Benford's Law: Distributuon of leading digits
1	30.10%
2	17.60%
3	12.50%
4	9.70%
5	7.90%
6	6.70%
7	5.80%
8	5.10%
9	4.60%

Detecting Fraud When working for the Brooklyn district attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford’s law, the check amounts appear to result from fraud. Use a 0.01 significance level to test for goodness-of-fit with Benford’s law. Does it appear that the checks are the result of fraud?

The frequencies of the different leading digits of the amounts of 784 checks are recorded.

Assume that random sampling is conducted.

Let O denote the observed frequencies of the leading digits.

The observed frequencies are noted below:

\(\begin{aligned}{c}{O_1} = 0\\{O_2} = 15\;\;\\{O_3} = 0\;\;\\{O_4} = 76\end{aligned}\)

\({O_5} = 479\)

\(\begin{aligned}{c}{O_6} = 183\\{O_7} = 8\;\;\\{O_8} = 23\;\;\\{O_9} = 0\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 0 + 15 + ...... + 0\\ = 784\end{aligned}\)

Let E denote the expected frequencies.

Let the expected proportion and expected frequencies of the ith digit as given by Benford’s law.

Since the expected values are larger than 5, the requirements of the test are met.

The null hypothesis for conducting the given test is as follows:

The observed frequencies are the same as the frequencies expected from Benford’s law.

The alternative hypothesis is as follows:

The observed frequencies are not the same as the frequencies expected from Benford’s law.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 235.984 + 109.6146 + ....... + 36.064\\ = 3650.251\end{aligned}\]

Thus,\({\chi ^2} = 3650.251\).

Let k be the number of digits, which is 9.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 9 - 1\\ = 8\end{aligned}\)

The chi-square table is used to obtain the critical value of\({\chi ^2}\)at\(\alpha = 0.01\)with 8 degrees of freedom is equal to 20.090.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 3650.251} \right)\\ = 0.000\end{aligned}\)

Since the test statistic value is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to conclude thatthe observed frequencies are not the same as the frequencies expected from Benford’s law.

Since the observed frequencies differ from the expected frequencies, the check amounts are a result of fraud.