Benford's law states that the probability distribution of the first digits of many items (e.g. populations and expenses) is not uniform, but has the probabilities shown in this table. Business expenses tend to follow Benford's Law, because there are generally more small expenses than large expenses. Perform a "Goodness of Fit" Chi-Squared hypothesis test $(\alpha=0.05)$ to see if these values are consistent with Benford's Law. If they are not consistent, it could imply there might be embezzelment. Complete this table. The sum of the observed frequencies is 77 \begin{tabular}{|l|l|l|l|} \hline $\mathbf{X}$ & \begin{tabular}{c} Observed \\ Frequency \\ (Counts) \end{tabular} & \begin{tabular}{c} Benford's \\ Law $P(X)$ \end{tabular} & \begin{tabular}{c} Expected \\ Frequency \\ (Counts) \end{tabular} \\ \hline 1 & 24 & .301 & \\ \hline 2 & 16 & .176 & \\ \hline 3 & 5 & .125 & \\ \hline 4 & 10 & .097 & \\ \hline 5 & 7 & .079 & \\ \hline 6 & 2 & .067 & \\ \hline 7 & 9 & .058 \\ \hline 8 & 2 & .051 \\ \hline 9 & 2 & .046 \\ \hline \end{tabular} Report all answers accurate to three decimal places. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.) \[ \chi^{2}= \] What is the P-value for this sample? (Report answer accurate to 3 decimal places.) $P$-value $=$ The P-value is... less than (or equal to) $\alpha$ greater than $\alpha$ This P-Value leads to a decision to... reject the null hypothesis fail to reject the null hypothesis As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that these expenses are consistent with Benford's Law.
Solution
Step 1 ：First, calculate the expected frequencies for each digit. The expected frequency is calculated by multiplying the total number of observations (77) by the probability of each digit according to Benford's Law.
Step 2 ：The expected frequencies are calculated as follows: \[23.177, 13.552, 9.625, 7.469, 6.083, 5.159, 4.466, 3.927, 3.542\]
Step 3 ：Next, calculate the chi-square test statistic using the formula: \[\chi^{2} = \sum \frac{(O_i - E_i)^2}{E_i}\] where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency.
Step 4 ：The chi-square test statistic is calculated as follows: \[\chi^{2} = 0.029 + 0.447 + 2.223 + 0.855 + 0.136 + 1.911 + 4.568 + 0.476 + 0.677 = 11.322\]
Step 5 ：The degrees of freedom for this test is 8 (9 categories - 1).
Step 6 ：The P-value is the probability that a chi-square statistic having 8 degrees of freedom is more extreme than 11.322. We can find this value using a chi-square distribution table or a calculator with a chi-square distribution function. The P-value is 0.185.
Step 7 ：Since the P-value (0.185) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Step 8 ：\(\boxed{\text{There is not sufficient evidence to warrant rejection of the claim that these expenses are consistent with Benford's Law.}}\)
