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%

Author’s Check Amounts 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?

The frequencies of the different leading digits of the amounts of 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:

\(\begin{aligned}{c}{O_1} = 83\\{O_2} = 58\;\;\\{O_3} = 27\;\;\\{O_4} = 21\end{aligned}\)

\({O_5} = 21\)

\(\begin{aligned}{c}{O_6} = 21\\{O_7} = 6\;\;\\{O_8} = 4\;\;\\{O_9} = 9\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 83 + 58 + ... + 9\\ = 250\end{aligned}\)

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 of leading digits are the same as the frequencies expected from Benford’s law.

The alternative hypothesis is as follows:

The observed frequencies of leading digits 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:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 0.798173 + 4.454545 + ....... + 0.543478\\ = 18.955\end{aligned}\)

Thus,\({\chi ^2} = 18.955\).

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

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

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

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

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 18.955} \right)\\ = 0.015\end{aligned}\).

Since the test statistic value is less than the critical value and the p-value is greater than 0.01, thenull hypothesis is failed to be rejected.

There is enough evidence to conclude that the observed frequencies of the leading digits are the same as the frequencies expected from Benford’s law.

The value of the chi-square test statistic is equal to 18.955.

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 8 degrees of freedom is equal to 15.507.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 18.955} \right)\\ = 0.015\end{aligned}\)

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected at a 0.05 level of significance.

Thus, it can be concluded that the observed frequencies of the leading digits from the amounts of checks are not the same as the expected frequencies using Benford’s law.

Thus, the conclusion changes as the level of significance changes from 0.01 to 0.05.