Chapter 11.1, Problem 11E

a.

To determine

To check: If the data is consistent with Benford’s law by conducting the significant test at % significance law.

Answer to Problem 11E

There is sufficient evidence to reject the claim of Benford’s claim.

Explanation of Solution

Given:

The data set is:

First digit (X)	Probability	Count
1	0.301	61
2	0.176	40
3	0.125	43
4	0.097	34
5	0.079	25
6	0.067	16
7	0.058	7
8	0.051	8
9	0.046	6

Formula used:

The formula to compute the chi-square statistic is:

  ${\chi }^2=\frac{{\displaystyle \sum {\left(\text{Observed - Expected}\right)}^2}}{\text{Expected}}$

Calculation:

The hypotheses are:

  $\begin{array}{l}{H}_0:p_1=0.301,p_2=0.176,p_3=\mathrm{0.125....}{p}_9=0.046\\ H_a:\text{At least one of the }{p}_i\text{ is different}\end{array}$

The calculation for test statistic is done as:

Observed value	Expected value	$\left(O-E\right)$	${\left(O-E\right)}^2$	${\left(O-E\right)}^2/E$
61	75.25	-14.25	20.30625	2.6985
50	44	6	36	0.8182
43	31.25	11.75	138.0625	4.418
34	24.25	9.75	95.0625	3.9201
25	19.75	5.25	27.5625	1.3956
16	16.75	-0.75	0.5625	0.0336
7	14.5	-7.5	56.25	3.8793
8	12.75	-4.75	22.5625	1.7696
6	11.5	-5.5	30.25	2.6304
				${\displaystyle \sum {\left(O-E\right)}^2/E}=21.5633$

The test statistic is:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum {\left(O-E\right)}^2/E}\\ =21.5633\end{array}$

The degree of freedom is calculated as:

  $\begin{array}{c}\text{Degree of freedom =Number of categories - 1}\\ =9-1\\ =8\end{array}$

The p-value using chi-square table at 8 degree of freedom is 0.005792

The p -value is below significance level. The null hypothesis gets rejected. Thus, there is sufficient evidence to reject the claim of Benford’s claim.

b.

To determine

To discuss: The Type I and Type II errors in the context of the problem. Also, explain the error that is more serious.

Answer to Problem 11E

Type I error.

Explanation of Solution

Type I error happenes if the null hypthesis is rejected even if it is correct. Here, type I error would occur if it is concluded that Benford’s law is not appropriate for the sample even it is.

On the other hand, type II error would occur if it is concluded that Benford’s law is appropriate for the sample even it is not.

Here, Type I error is more worst because it will accuse an innocent company to commit fraud.