Q7 **Understanding Benford's Law and Its Application****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.### Hypothesis Testing using Chi-Squared TestWe will perform a "Goodness of Fit" Chi-Squared hypothesis test (α = 0.05) to see if these values are consistent with Benford's Law. If they are not consistent, it might indicate possible discrepancies such as embezzlement.### Data TableBelow is a table illustrating the observed frequencies, expected probabilities according to Benford’s Law, and the expected frequency counts.| X (First Digit) | Observed Frequency (Counts) | Benford’s Law P(X) | Expected Frequency (Counts) ||-----------------|-----------------------------|--------------------|-----------------------------|| 1 | 46 | 0.301 | || 2 | 23 | 0.176 | || 3 | 9 | 0.125 | || 4 | 13 | 0.097 | || 5 | 7 | 0.079 | || 6 | 7 | 0.067 | || 7 | 9 | 0.058 | || 8 | 6 | 0.051 | || 9 | 4 | 0.046 | |*Note: The sum of the observed frequencies is 124.*### Hypotheses- **Null Hypothesis (H₀):** The observed distribution follows Benford's Law.- **Alternative Hypothesis (H₁):** At least one of the observed frequencies differs from the expected according to Benford's Law.### Calculation of Expected FrequenciesTo complete the table, multiply each probability by the total number of observations (124):- Expected frequency for first digit 1: \( 0.301 \times 124 \)- Expected frequency for first digit 2: \( 0.176 \times 124 \)- Expected frequency for first digit 3: \( 0.125 ### Conclusion:- ☐ There is sufficient evidence to warrant rejection of the claim that these values are consistent with Benford's Law.- ☐ There is not sufficient evidence to warrant rejection of the claim that these values are consistent with Benford's Law.- ☐ The sample data supports the claim that these values are consistent with Benford's Law.- ☐ There is not sufficient data to support the claim that these values are consistent with Benford's Law.### Instructions:Enter the critical value, along with the significance level and degrees of freedom \( \chi^2 (\alpha, df) \) below the graph. *(Graph is for illustration only. No need to shade.)*### Graph:#### χ²-Distribution![χ²-Distribution graph illustration](link_to_image)**Description of the Graph:**- The x-axis represents the values of \( \chi^2 \) (Chi-Square).- The y-axis represents the probability.- The curve illustrates a typical Chi-Square distribution which starts at 0 on the y-axis, rises to a peak, and then gradually declines towards the x-axis.### Input Fields:\[ \chi^2 \, ( \, \, ) \, = \, \, \, \]This text and diagram explanation assist in understanding the statistical conclusion related to Benford's Law, Chi-Square distributions, and the process of determining critical values.

Name: Q7 **Understanding Benford's Law and Its Application****Benford's law*
Uploaded: 2024-03-20T06:47:08.000Z
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Description: Q7 **Understanding Benford's Law and Its Application****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. Busines

Given: X observed frequency p(x) 1 46 0.301 2 23 0.176 3 9 0.125 4 13 0.097 5 7…

Math

Statistics

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 (a = 0.05) to see if these values are consistent with Benford's Law. If they are not consistent, it there might be embezzelment. Complete this table. Observed X Frequency (Counts) Benford's Law P(X) Expected Frequency (Counts) 1 46 .301 2 23 .176 3 9 .125 4 13 .097 5 7 .079 6 7 .067 79 .058 86 .051 9 4 .046 The sum of the observed frequencies is 124. Но: рі - .301, Р2 — .176, рз в .125, ра .097, рs - .079, рs 3 .067, р7 .058, ps- .051, ро 3.046 H1: at least one is different Original Claim: H, OH, Test Statistic = (Round to three decimal places.) P-value = (Round to four decimal places.) Decision: o reject the null hypothesis fail to reject the null hypothesis O Accept the null hypothesis