Question

n investigator analyzed the leading digits from 799 checks issued by seven suspect companies. The frequencies were found to be 235​, 148​, 96​, 90​, 68​, 47​, 43​, 38​, and 34​, 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 0.01 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 235 148 96 90 68 47 43 38 34 ​Benford's Law: Distribution of Leading Digits ​30.1% ​17.6% ​12.5% ​9.7% ​7.9% ​6.7% ​5.8% ​5.1% ​4.6%

Calculate the test​ statistic,

chi squaredχ2.

chi squaredχ2equals=nothing

​(Round to three decimal places as​ needed.)

Calculate the​ P-value.

​P-valueequals=nothing

​(Round to four decimal places as​ needed.)

State the conclusion.

▼

Reject

Do not reject

Upper H 0H0.

There

▼

is not

is

sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to​ Benford's law. It

▼

does appear

does not appear

that the checks are the result of fraud.

Answer 1

We want to test whether the observed values follow the Benford's law that is to test if the distribution is good fit or not. Therefore we use to chi-square goodness of fit test.

We have been given the observed values and the expected proportions.

Using themwe find the expected frequnecy.

Expected freq = Exp proportion * Total freq

Number	Observed	Exp proportion	Expected	Oi-Ei
1	235	30.1%	240.499	-5.499	0.1257
2	148	17.6%	140.624	7.376	0.3869
3	96	12.5%	99.875	-3.875	0.1503
4	90	9.7%	77.503	12.497	2.0151
5	68	7.9%	63.121	4.879	0.3771
6	47	6.7%	53.533	-6.533	0.7973
7	43	5.8%	46.342	-3.342	0.2410
8	38	5.1%	40.749	-2.749	0.1855
9	34	4.6%	36.754	-2.754	0.2064
	799	100.0%			4.4853

Test

Null: The observed and expected values are same and so the checks are not fraud.

Alternative: The observed and expected values are not same and so the checks are fraud.

Chi -square test STat =

Test Stat = 4.4853

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

=P( > 4.49) .................using chi-square tables

p-value = 0.8109

Level of significance = 0.05

Since p-value > 0.05

• State the conclusion.

Do not reject

• There is not sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to​ Benford's law. It
• does not appear that the checks are the result of fraud.