ou might think that if you looked at the first digit in randomly
selected numbers that the distribution would be uniform. Actually,
it is not! Simon Newcomb and later Frank Benford both discovered
that the digits occur according to the following distribution:
(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)(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)
The IRS currently uses Benford's Law to detect fraudulent tax data.
Suppose you work for the IRS and are investigating an individual
suspected of embezzling. The first digit of 166 checks to a
supposed company are as follows:
Digit | Observed Frequency |
---|---|
1 | 40 |
2 | 20 |
3 | 14 |
4 | 17 |
5 | 19 |
6 | 14 |
7 | 8 |
8 | 17 |
9 | 17 |
a. State the appropriate null and alternative hypotheses for this
test.
b. Explain why ?=0.01?=0.01 is an appropriate choice for the level
of significance in this situation.
c. What is the P-Value? Report answer to 4 decimal places.
d. What is your decision?
Write a statement to the law enforcement officials that will use it
to decide whether to pursue the case further or not. Structure your
essay as follows: Given a brief explanation of what a Goodness of
Fit test is. Explain why a Goodness of Fit test should be applied
in this situation. State the hypotheses for this situation
Interpret the answer to part c Use the answer to part c to justify
the decision in part d Use the decision in part d to make a
conclusion about whether the individual is likely to have
embezzled. Use this to then tell the law enforcement officials
whether they should pursue the case or not.
c)
Oi | Ei | (Oi-Ei)^2/Ei | |
0.301 | 40 | 47.257 | 1.114417949 |
0.176 | 20 | 27.632 | 2.10796989 |
0.125 | 14 | 19.625 | 1.612261146 |
0.097 | 17 | 15.229 | 0.205951868 |
0.079 | 19 | 12.403 | 3.508861485 |
0.067 | 14 | 10.519 | 1.1519499 |
0.058 | 8 | 9.106 | 0.134332967 |
0.051 | 17 | 8.007 | 10.10041826 |
0.046 | 17 | 7.222 | 13.2386159 |
1 | 166 | 157 | 33.17477936 |
df= n-1= 9-1 = 8
p-value =
=1 - CHISQ.DIST(36.142506,8,1)
= 0.00001
d)
since p-value < 0.05
we reject the null hypothesis
we conclude that the individual is likely to have embezzled.
the law enforcement officials should pursue the case
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