Question

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 Computer Files The author recorded the leading digits of the sizes of the electronic document files for the current edition of this book. The leading digits have frequencies of 55, 25, 17, 24, 18, 12, 12, 3, and 4 (corresponding to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively). Using a 0.05 significance level, test for goodness-of-fit with Benford’s law.

Short answer

There is not enough evidence to conclude that the observed frequencies of the leading digits of the sizes of the electronic document files are not the same as the frequencies expected from Benford’s law.

Step-by-step solution

Given information

The frequencies of the different leading digits from IRS tax files are recorded.

Step 2:Check the requirements

Assume that random sampling is conducted.

Let O denote the observed frequencies of the leading digits.

The observed frequencies are noted below:

\(\begin{aligned}{c}{O_1} = 55\\{O_2} = 25\\{O_3} = 17\;\;\\{O_4} = 24\end{aligned}\)

\({O_5} = 18\)

\(\begin{aligned}{c}{O_6} = 12\\{O_7} = 12\;\;\\{O_8} = 3\;\;\\{O_9} = 4\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 55 + 25 + ... + 4\\ = 170\end{aligned}\)

Let E denote the expected frequencies.

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

Leading Digits	Benford's Law: Distribution of leading digits	Proportions	Expected Frequencies
1	30.10%	\(\begin{aligned}{c}{p_1} = \frac{{30.1}}{{100}}\\ = 0.301\end{aligned}\)	\(\begin{aligned}{c}{E_1} = n{p_1}\\ = 170\left( {0.301} \right)\\ = 51.17\end{aligned}\)
2	17.60%	\(\begin{aligned}{c}{p_2} = \frac{{17.6}}{{100}}\\ = 0.176\end{aligned}\)	\(\begin{aligned}{c}{E_2} = n{p_2}\\ = 170\left( {0.176} \right)\\ = 29.90\end{aligned}\)
3	12.50%	\(\begin{aligned}{c}{p_3} = \frac{{12.5}}{{100}}\\ = 0.125\end{aligned}\)	\(\begin{aligned}{c}{E_3} = n{p_3}\\ = 170\left( {0.125} \right)\\ = 21.25\end{aligned}\)
4	9.70%	\(\begin{aligned}{c}{p_4} = \frac{{9.7}}{{100}}\\ = 0.097\end{aligned}\)	\(\begin{aligned}{c}{E_4} = n{p_4}\\ = 170\left( {0.097} \right)\\ = 16.49\end{aligned}\)
5	7.90%	\(\begin{aligned}{c}{p_5} = \frac{{7.9}}{{100}}\\ = 0.079\end{aligned}\)	\(\begin{aligned}{c}{E_5} = n{p_5}\\ = 170\left( {0.079} \right)\\ = 13.43\end{aligned}\)
6	6.70%	\(\begin{aligned}{c}{p_6} = \frac{{6.7}}{{100}}\\ = 0.067\end{aligned}\)	\(\begin{aligned}{c}{E_6} = n{p_6}\\ = 170\left( {0.067} \right)\\ = 11.39\end{aligned}\)
7	5.80%	\(\begin{aligned}{c}{p_7} = \frac{{5.8}}{{100}}\\ = 0.058\end{aligned}\)	\(\begin{aligned}{c}{E_7} = n{p_7}\\ = 170\left( {0.058} \right)\\ = 9.86\end{aligned}\)
8	5.10%	\(\begin{aligned}{c}{p_8} = \frac{{5.1}}{{100}}\\ = 0.051\end{aligned}\)	\(\begin{aligned}{c}{E_8} = n{p_8}\\ = 170\left( {0.051} \right)\\ = 8.67\end{aligned}\)
9	4.60%	\(\begin{aligned}{c}{p_9} = \frac{{4.6}}{{100}}\\ = 0.046\end{aligned}\)	\(\begin{aligned}{c}{E_9} = n{p_9}\\ = 170\left( {0.046} \right)\\ = 7.82\end{aligned}\)

As all the expected values are higher than 5, the requirements of the test are satisfied.

State the hypotheses

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 test is right-tailed.

If the absolute value of the test statistic is greater than the critical value, the null hypothesis is rejected.

Conduct the hypothesis test

The table below shows the necessary calculations:

Leading Digits	O	E	\(\left( {O - E} \right)\)	\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)
1	55	51.17	3.83	0.286670
2	25	29.92	-4.92	0.809037
3	17	21.25	-4.25	0.850000
4	24	16.49	7.51	3.420261
5	18	13.43	4.57	1.555093
6	12	11.39	0.61	0.032669
7	12	9.86	2.14	0.464462
8	3	8.67	-5.67	3.708062
9	4	7.82	-3.82	1.866036

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\\ = 0.28667 + 0.809037 + ... + 1.866036\\ = 12.992\end{aligned}\)

Thus,\({\chi ^2} = 12.992\).

Let k be the number of digits, equal to 9.

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

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

State the conclusion

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 8 degrees of freedom is equal to 15.507, taken from the chi-square table.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 12.992} \right)\\ = 0.112\end{aligned}\)

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is not enough evidence to conclude that the observed frequencies of the leading digits of the sizes of the electronic document files are not the same as the frequencies expected from Benford’s law.