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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.

 Apr 4, 2015

Best Answer 

 #1
avatar+21951 
+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.

 Apr 5, 2015
 #1
avatar+21951 
+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+110208 
0

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

 Apr 5, 2015

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