Question

Cybersecurity The table below lists leading digits of 317 inter-arrival Internet traffic times for a computer, along with the frequencies of leading digits expected with Benford’s law (from Table 11-1 in the Chapter Problem).

a. Identify the notation used for observed and expected values.

b. Identify the observed and expected values for the leading digit of 2.

c. Use the results from part (b) to find the contribution to the${\chi }^2$test statistic from the category representing the leading digit of 2.

Leading Digit	1	2	3	4	5	6	7	8	9
Benford’s Law	30.1%	17.6%	12.5%	9.7%	7.9%	6.7%	5.8%	5.1%	4.6%
Leading Digits of Inter-Arrival Traffic Times	76	62	29	33	19	27	28	21	22

Short answer

a.The notation used for the observed frequencies is O and the notation used for the expected frequencies is E.

b.The observed frequency for the leading digit of 2 is equal to 62 and the expected frequency of the leading digit of 2 is equal to 55.792.

c. The value of$\frac{{\left(O-E\right)}^2}{E}$is equal to 0.691.

Step-by-step solution

Given information

The observed frequencies and the expected frequencies of the leading digits of inter-arrival traffic times are tabulated.

Notation for observed and expected frequency

a.

The observed frequencies are represented by O.

The expected frequencies are represented by E.

Identify the observed and expected frequency for the leading digit 2

b.

The observed frequency (O) corresponding to the leading digit of 2 is equal to 62.

The formula to compute the expected frequency is as follows:

$E=np$where

n is the sum of all observed frequencies

p is the probability of occurrence of that outcome.

Here, the sum of all observed frequencies is computed below:

$\begin{array}{r}cn=76+62+.....+22\\ =317\end{array}$

The value of p for the leading digit of 2 is equal to:

$\begin{array}{r}cp=17.6\mathrm{\%}\\ =\frac{17.6}{100}\\ =0.176\end{array}$

Thus, the expected frequency of the leading digit of 2 is computed below:

$\begin{array}{r}cE=np\\ =317\left(0.176\right)\\ =55.792\end{array}$

Therefore, the expected frequency of the leading digit of 2 is equal to 55.792.

Contribution to the test statistic

c.

The contribution to thechi-squaretest statistic from the category representing the leading digit of 2 is computed below:

$\begin{array}{r}c\frac{{\left(O-E\right)}^2}{E}=\frac{{\left(62-55.792\right)}^2}{55.792}\\ =0.6907\\ \approx 0.691\end{array}$

Thus, the contribution to the chi-squaretest statistic from the category representing the leading digit of 2 is equal to 0.691.