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You might think that if you looked at the first digit in randomly selected numbe

ID: 3152238 • Letter: Y

Question

You 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 155 checks to a supposed company are as follows:



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
P-Value =  


d. What is your decision?

Reject the Null Hypothesis

Fail to reject the Null Hypothesis



e. Write a statement to the law enforcement officials that will use it to decide whether to pursue the case further or not. Include the following: explanation of the hypotheses, interpretation of part c, why this leads to the decision in part d, and what your recommendation is for the case.

Digit Observed
Frequency 1 45 2 25 3 28 4 9 5 22 6 4 7 13 8 0 9 9

Explanation / Answer

answer a) Null hypothesis Ho: probability of each each digit (1 to 9) is equal

alternative hypotheis H1: probability at least two digits are not equal

answer b) based on tradition, generally alpha is taken as 0.05 in social scienece data

but delicated data like revenue, corruption we go for higher level of confidence level and corresponding lower value of alpha. That is why alpha=0.01 is appropriate in this case

answer c) here we use goodness of fit test using chi-square

and chi-square=sum((O-E)2/E) with k-1 degree of freedom

chi-square=92.6452 with 8 df and corresponding p-value <0.001

answer p-value <0.001

answer d) since p-value is less than alpha so we reject null hypothesis

answer e) since probability of each each digit (1 to 9) is not equal which was tested using chi-square goodness of fit and it is found  embezzling and should be further investivate the case in details.

Digit Observed(O) Expected(E) Frequency Frequency O-E (O-E)2 (O-E)2/E 1 45 17.22222222 27.77778 771.6049 44.80287 2 25 17.22222222 7.777778 60.49383 3.512545 3 28 17.22222222 10.77778 116.1605 6.744803 4 9 17.22222222 -8.22222 67.60494 3.925448 5 22 17.22222222 4.777778 22.82716 1.325448 6 4 17.22222222 -13.2222 174.8272 10.15125 7 13 17.22222222 -4.22222 17.82716 1.035125 8 0 17.22222222 -17.2222 296.6049 17.22222 9 9 17.22222222 -8.22222 67.60494 3.925448 Total= 155 92.64516
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