Question

An investigator analyzed the leading digits from 787 checks issued by seven suspect companies. The frequencies were found to be 4​, 11​, 2​, 72​, 371​, 281​, 7​, 16​, and 23​, 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 shown​ below, the check amounts appear to result from fraud. Use a 0.025 significance level to test for​ goodness-of-fit with​ Benford's law. Does it appear that the checks are the result of​ fraud?

Leading Digit:      1    2    3 4 5 6    7 8 9
Actual Frequency:      4 11    2    72    371 281 7    16    23

Benford's Law:      30.1% 17.6% 12.5%    9.7%    7.9% ​ 6.7%    ​5.8% ​ 5.1%    ​4.6%

Determine the null and alternative hypotheses.

Ho​: (1)_________________    H1​: (2)_________________

Calculate the test​ statistic, χ2.

χ2 = _______________​

(Round to three decimal places as​ needed.)

Calculate the​ P-value.

P-value = _______________​

(Round to four decimal places as​ needed.)

Answer 1

We are testing the given observed values follow the Benford's Law. That is we are testing if a distribution is good fit or not. For this we will conduct a chi-square goodness of fit test.

We need the expected values as well. Since we have been given the expected proportions

Expected value = expected proportion * Observed total

LD	Actual / obs	Proportion	Expected	O - E	(O - E)2
1	4	0.301	236.887	-232.887	54236.355	228.955
2	11	0.176	138.512	-127.512	16259.310	117.386
3	2	0.125	98.375	-96.375	9288.141	94.416
4	72	0.097	76.339	-4.339	18.827	0.247
5	371	0.079	62.173	308.827	95374.116	1534.012
6	281	0.067	52.729	228.271	52107.649	988.216
7	7	0.058	45.646	-38.646	1493.513	32.719
8	16	0.051	40.137	-24.137	582.595	14.515
9	23	0.046	36.202	-13.202	174.293	4.814
Total	787		787			3015.279

Determine the null and alternative hypotheses.

Ho​: (1)The amounts follow Benford's Law. That is the checks are not faulty.

H1​: (2)The amounts do not follow Benford's Law. That is the checks are faulty.

Calculate the test​ statistic, χ2.

Chi -square test STat =

Test Stat = 3015.279

Calculate the​ P-value.

P-value =P( > Test stat) ....................df = n-1

= P(₈ > 3015.279)

p-value = 0.000

Since p-value < 0.025

We reject the null hypothesis at 2.5%. There is sufficient evidence to conclude that the amounts do not follow Benford's Law. That is the checks are faulty.