An investigator analyzed the leading digits of the amounts from 200 checks issued by three suspect companies. the frequencies were found to be 68, 40, 18, 19, 8, 20, 6, 9, 12 and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. if the observed frequencies are substantially different from the frequencies expected with benford's law, the check amounts appear to be the result of fraud. use a 0.05 significance level to test for goodness-of-fit with benford's law. 1. calculate the χ2 test statistic.

The observed p-value is 0.069669438 which does not indicate fraud to the 0.05 significance level specified. Benford's law was the observation that the leading digits of numbers have the percentage change of appearing proportional to the logarithm of the number. It was first noticed by Simon Newcomb in 1881 when he noticed that earlier pages in a book of logarithms were much more worn that later pages. Newcomb proposed the law that the probability of a single digit N being the 1st digit of a number was log(N+1) - log(N). Frank Benford noticed the phenomenon in 1938 and tested it against multiple domains of data and is credited with the law. So with 200 data points, the expected distribution would be 60, 35, 25, 19, 16, 13, 12, 10, and 9. Let's calculate the Chi-Square value. The formula is χ2 = sum of (Oi - Ei)^2/Ei where Oi is the ith observation and Ei is the expected observation. 1: (68 - 200*(log(1+1) - log(1)))^2/(200*(log(1+1) - log(1))) = (68-60.206)^2/60.206 = 1.008976687 2: (40 - 200*(log(2+1) - log(2)))^2/(200*(log(2+1) - log(2))) = 0.649240509 3: (18 - 200*(log(3+1) - log(3)))^2/(200*(log(3+1) - log(3))) = 1.954102225 4: (19 - 200*(log(4+1) - log(4)))^2/(200*(log(4+1) - log(4))) = 0.007528943 5: (8 - 200*(log(5+1) - log(5)))^2/(200*(log(5+1) - log(5))) = 3.877610213 6: (20 - 200*(log(6+1) - log(6)))^2/(200*(log(6+1) - log(6))) = 3.263830041 7: (6 - 200*(log(7+1) - log(7)))^2/(200*(log(7+1) - log(7))) = 2.70226863 8: (9 - 200*(log(8+1) - log(8)))^2/(200*(log(8+1) - log(8))) = 0.148002603 9: (12 - 200*(log(9+1) - log(9)))^2/(200*(log(9+1) - log(9))) = 0.886626747 The sum of the values is 14.4981866 Since we have 9 different values being tested, we only have 8 degrees of freedom. So look up in a chi-square table for the value 14.498 with 8 degrees of freedom, getting: 0.069669438. This value indicates that there's about a 6.9% chance that the observed p-value would have occurred randomly which is above the 0.05 significance value if just barely. So the chi-square test for this data indicates that fraud did not happen.