91Ó°ÊÓ

According to Benford's law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below.Test for goodness-of-fit with the distribution described by Benford's law. $$\begin{array}{l|c|c|c|c|c|c|c|c|c} \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \begin{array}{l} \text { Benford's Law: Distribution } \\ \text { of Leading Digits } \end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\ \hline \end{array}$$ The author recorded the leading digits of the sizes of the clectronic document files for the current edition of this book. The leading digits have frequencies of \(55,25,17,24,18,12,12,3,\) and 4 (corresponding to the leading digits of 1,2,3,4,5,6,7,8 and \(9,\) respectively). Using a 0.05 significance level, test for goodness-of-fit with Benford's law.

Step by step solution

01

- Define the Hypotheses

Set up the null and alternative hypotheses. The null hypothesis \(H_0\) states that the leading digits follow Benford's Law. The alternative hypothesis \(H_1\) states that the leading digits do not follow Benford's Law.

02

- Calculate Expected Frequencies

Using the sample size of the leading digits (in this case, the sum of the frequencies of the digits), compute the expected frequencies by multiplying the sample size by the Benford's Law proportions. The sample size \(n = 55 + 25 + 17 + 24 + 18 + 12 + 12 + 3 + 4 = 170\). For each digit, \[ \text{Expected Frequency} = \text{Benford's Law percentage} \times \text{Total sample size} \]

03

- Perform Calculations for Each Digit

Calculate the expected frequency for each leading digit: \(1: 170 \times 0.301 = 51.17\), \(2: 170 \times 0.176 = 29.92\), \(3: 170 \times 0.125 = 21.25\), \(4: 170 \times 0.097 = 16.49\), \(5: 170 \times 0.079 = 13.43\), \(6: 170 \times 0.067 = 11.39\), \(7: 170 \times 0.058 = 9.86\), \(8: 170 \times 0.051 = 8.67\), \(9: 170 \times 0.046 = 7.82\).

04

- Conduct Chi-Square Test

Use the Chi-Square test formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies. Calculate it for each digit: \(\chi^2 = \frac{(55 - 51.17)^2}{51.17} + \frac{(25 - 29.92)^2}{29.92} + ... + \frac{(4 - 7.82)^2}{7.82} \)

05

- Find Chi-Square Critical Value

Determine the chi-square critical value at the 0.05 significance level with degrees of freedom \(df = \text{number of categories} - 1 = 9 - 1 = 8\). Look up the critical value in a chi-square distribution table, which is 15.507.

06

- Compare and Conclude

Compare the computed chi-square statistic with the critical value. If \(\chi^2\) is less than or equal to 15.507, we fail to reject the null hypothesis. Otherwise, we reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Goodness-of-Fit Test

The Chi-Square Goodness-of-Fit Test is a statistical hypothesis test used to determine if a sample data matches a population with a specific distribution. In our case, we want to check if the leading digits of the electronic document file sizes follow Benford's Law.

The test involves several steps: calculating the expected frequencies based on Benford's Law, comparing these expected frequencies with the observed frequencies from the sample data, and then computing the chi-square statistic using the formula:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
Here, \( O_i \) represents the observed frequencies, and \( E_i \) the expected frequencies. The result is then compared against a critical value from the chi-square distribution table to draw conclusions about our hypotheses.

Null and Alternative Hypotheses

Hypothesis testing begins with the formulation of two competing statements: the null hypothesis and the alternative hypothesis.

The **null hypothesis** \( H_0 \) assumes that there is no significant difference between the observed data and the expected data according to a specific distribution. For our exercise, this would mean that the leading digits follow Benford's Law.

Conversely, the **alternative hypothesis** \( H_1 \) suggests that there is a significant difference, implying that the leading digits do not follow Benford's Law. The goal of the test is to collect evidence to either reject or fail to reject the null hypothesis.

Expected and Observed Frequencies

Understanding expected and observed frequencies is crucial in carrying out the Chi-Square Goodness-of-Fit Test.

**Observed frequencies** are the actual counts we obtain from our sample data. In our context, this means counting how many times each digit (1 through 9) appears as the leading digit in the document file sizes.
For instance:

• Digit 1: 55
• Digit 2: 25
• ... and so on

**Expected frequencies** are the counts we expect to find if the data perfectly follows the theoretical distribution, in this case, Benford's Law. These are calculated by multiplying the total sample size by the percentage distribution provided by Benford's Law.
For digit 1:
\[ \text{Expected Frequency} = \text{170} \times 0.301 = 51.17 \]

Significance Level in Hypothesis Testing

The significance level, denoted by \( \alpha \), is a threshold used to determine whether the observed data is significantly different from the expected data under the null hypothesis.

Commonly set at 0.05 (5%), the significance level represents the probability of rejecting the null hypothesis when it is actually true.

In our test:

• If the chi-square statistic is less than or equal to the critical value at \( \alpha = 0.05 \), we **fail to reject the null hypothesis**, concluding that the leading digits conform to Benford's Law.
• If it exceeds the critical value, we **reject the null hypothesis**, suggesting that the leading digits do not follow Benford's Law.

For this exercise:

• Degrees of freedom: 8 (number of categories minus one)
• Critical value at \( \alpha = 0.05\) is 15.507