91Ó°ÊÓ

Set up the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). The null hypothesis states that the leading digits follow Benford's law, and the alternative hypothesis states that they do not follow Benford's law.\[ H_0: \text{The leading digits follow Benford's law.} \]\[ H_a: \text{The leading digits do not follow Benford's law.} \]

Benford's law gives the expected proportion for each leading digit. Calculate the expected frequency for each digit by multiplying the total number of observations by the proportion given by Benford's law.The expected proportions are:\[ P(1) = \log_{10}(1 + \frac{1}{1}) \approx 0.301 \]\[ P(2) = \log_{10}(1 + \frac{1}{2}) \approx 0.176 \]\[ P(3) = \log_{10}(1 + \frac{1}{3}) \approx 0.125 \]\[ P(4) = \log_{10}(1 + \frac{1}{4}) \approx 0.097 \]\[ P(5) = \log_{10}(1 + \frac{1}{5}) \approx 0.079 \]\[ P(6) = \log_{10}(1 + \frac{1}{6}) \approx 0.067 \]\[ P(7) = \log_{10}(1 + \frac{1}{7}) \approx 0.058 \]\[ P(8) = \log_{10}(1 + \frac{1}{8}) \approx 0.051 \]\[ P(9) = \log_{10}(1 + \frac{1}{9}) \approx 0.046 \]Total observations: \ 55 + 25 + 17 + 24 + 18 + 12 + 12 + 3 + 4 = 170.Expected frequencies are calculated as: \(E_i = 170 \times P(i)\).\[ E(1) = 170 \times 0.301 \approx 51.17 \]\[ E(2) = 170 \times 0.176 \approx 29.92 \]\[ E(3) = 170 \times 0.125 \approx 21.25 \]\[ E(4) = 170 \times 0.097 \approx 16.49 \]\[ E(5) = 170 \times 0.079 \approx 13.43 \]\[ E(6) = 170 \times 0.067 \approx 11.39 \]\[ E(7) = 170 \times 0.058 \approx 9.86 \]\[ E(8) = 170 \times 0.051 \approx 8.67 \]\[ E(9) = 170 \times 0.046 \approx 7.82 \]

Use the chi-square goodness-of-fit test formula:\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Substitute the observed (O) and expected (E) frequencies:\[ \chi^2 = \frac{(55 - 51.17)^2}{51.17} + \frac{(25 - 29.92)^2}{29.92} + \frac{(17 - 21.25)^2}{21.25} + \frac{(24 - 16.49)^2}{16.49} + \frac{(18 - 13.43)^2}{13.43} + \frac{(12 - 11.39)^2}{11.39} + \frac{(12 - 9.86)^2}{9.86} + \frac{(3 - 8.67)^2}{8.67} + \frac{(4 - 7.82)^2}{7.82} \approx 14.56 \]

The degrees of freedom (df) for the chi-square test is given by:\[ df = k - 1 = 9 - 1 = 8 \]

Using a significance level of \(\alpha = 0.05\), look up the critical value in a chi-square distribution table for 8 degrees of freedom.The critical value is approximately \(15.51\).Compare the test statistic to the critical value:\[ \chi^2 = 14.56 \text{ and critical value} = 15.51 \]Since \(\chi^2 = 14.56\) is less than the critical value \(15.51\), we fail to reject the null hypothesis.