+0  
 
0
2499
2
avatar

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

Best Answer 

 #1
avatar+17705 
+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
Sort: 

2+0 Answers

 #1
avatar+17705 
+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
 #2
avatar+91412 
0

That  is a great answer Geno.  I had forgotten that that is how it is done.  

Melody  Apr 5, 2015

17 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details