The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer.
Digit, Probability
1, 0.301
2, 0.176
3, 0.125
4, 0.097
5, 0.079
6, 0.067
7, 0.058
8, 0.051
9, 0.046
Fradulent Checks
Digit, Frequency
1, 36
2, 32
3, 45
4, 20
5, 24
6, 36
7, 15
8, 16
9, 7
Complete parts (a) and (b).
(a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?<br />
Yes or No
Based on the results of part (a), could one think that the employe is guilty of embezzlement?
Yes or No
Show frequency percentages
Digit Fraud Probability Benford Probability
1 0.156 0.301
2 0.139 0.176
3 0.195 0.125
4 0.087 0.097
5 0.104 0.079
6 0.156 0.067
7 0.065 0.058
8 0.069 0.051
9 0.03 0.046
Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277
Critical Value Excel: =CHIINV(0.95,8) = 2.733
Since test stat is less than critical value, we cannot reject, so YES, it does obey Benford's Law and NO, there is not enough evidence to suggest the employee is guilty of embezzlement.
Digit, Probability
1, 0.301
2, 0.176
3, 0.125
4, 0.097
5, 0.079
6, 0.067
7, 0.058
8, 0.051
9, 0.046
Fradulent Checks
Digit, Frequency
1, 36
2, 32
3, 45
4, 20
5, 24
6, 36
7, 15
8, 16
9, 7
Complete parts (a) and (b).
(a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?<br />
Yes or No
Based on the results of part (a), could one think that the employe is guilty of embezzlement?
Yes or No
Show frequency percentages
Digit Fraud Probability Benford Probability
1 0.156 0.301
2 0.139 0.176
3 0.195 0.125
4 0.087 0.097
5 0.104 0.079
6 0.156 0.067
7 0.065 0.058
8 0.069 0.051
9 0.03 0.046
Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277
Critical Value Excel: =CHIINV(0.95,8) = 2.733
Since test stat is less than critical value, we cannot reject, so YES, it does obey Benford's Law and NO, there is not enough evidence to suggest the employee is guilty of embezzlement.