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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鈥檚 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鈥檚 law.

The alternative hypothesis is as follows:

The observed frequencies of leading digits are not the same as the frequencies expected from Benford鈥檚 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鈥檚 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鈥檚 law.

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