According to Benford's law (Example 10.6, page 227) the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is 0.477. You examine 90 invoices from a vendor and find that 29 have first digits 1 or 2. If Benford's law holds, the count of 1s and 2s will have the binomial distribution with n=90 and p=0.477. Too few 1s and 2s suggests fraud.

Guest Apr 4, 2015

#1**+10 **

For a binomial distribution with an estimated success rate of 0.477 from a sample of 90:

μ = np = 90(0.477) = 42.93

σ = √(npq) = √[(90)(0.477)(0.523)] = √(22.) = 4.74

Using the normal approximation to the binomial with a continuity correction:

lower z-score = (28.5 - 42.93)/4.74 = -3.044

upper z-score = (29.5 - 42.93)/4.74 = -2.833

Probability of existing between those two z-scores: 0.0011

A P-score that low allows one to reject the claim that this data matches Benford's law.

geno3141 Apr 5, 2015

#1**+10 **

Best Answer

For a binomial distribution with an estimated success rate of 0.477 from a sample of 90:

μ = np = 90(0.477) = 42.93

σ = √(npq) = √[(90)(0.477)(0.523)] = √(22.) = 4.74

Using the normal approximation to the binomial with a continuity correction:

lower z-score = (28.5 - 42.93)/4.74 = -3.044

upper z-score = (29.5 - 42.93)/4.74 = -2.833

Probability of existing between those two z-scores: 0.0011

A P-score that low allows one to reject the claim that this data matches Benford's law.

geno3141 Apr 5, 2015