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 177 checks to a supposed company are as follows:

Digit	1	2	3	4	5	6	7	8	9
Observed Frequency	53	20	19	21	11	16	17	9	11

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. 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.

Answer 1

Chi square goodness of fit test:

Hypothesis:

H0: Observed frequency is equal to expected value

Ha: Observed freqency is different from expected value

b) Significant value= 0.01, it reduces the occuring of false poisitve(type 1 error)

The degree of freeodm= n-1=8

c)

Digit	Ob Frequency	Probability	Expected	(O-E)	(O-E)^2	(O-E)^2/E
1	53	0.301	53.277	-0.277	0.076729	0.00144
2	20	0.176	31.152	-11.152	124.3671	3.992267
3	19	0.125	22.125	-3.125	9.765625	0.441384
4	21	0.097	17.169	3.831	14.67656	0.854829
5	11	0.079	13.983	-2.983	8.898289	0.636365
6	16	0.067	11.859	4.141	17.14788	1.44598
7	17	0.058	10.266	6.734	45.34676	4.417179
8	9	0.051	9.027	-0.027	0.000729	8.08E-05
9	11	0.046	8.142	2.858	8.168164	1.003213
Sum	177				Sum	12.79274

P-value: 0.119

d) Fail to reject null hypothesis

e) There is sufficient evidence to support that the observed values equal to the expectred values and the law enforcement officials allowed to use this report.