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Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.2 - Problem 24bsc
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.2 - Problem 24bsc

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# Solution: Benford’s Law. According to Benford’s law, a

ISBN: 9780321836960 18

## Solution for problem 24BSC Chapter 11.2

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition

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Problem 24BSC

Benford’s Law. 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. In Exercise?, ?test for goodness-offit with Benford’s law. L e a d i n g D i g i t B e n f o r d ’ s L a w : D i s t r i b u t i o n o f L e a d i n g D i g i t s Author’s Computer Files? The author recorded the leading digits of the sizes of the files stored on his computer, and the leading digits have frequencies of 45, 32, 18, 12, 9, 3, 13, 9, and 9 (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:

Solution 24BSC Step 1 Given, the author recorded the leading digits of the sizes of the files stored on his computer, and the leading digits have frequencies of 45, 32, 18, 12, 9, 3, 13, 9, and 9 (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. We are testing Chi-square test (using EXCEL) as shown below Goodness of Fit Test expecte (Oi Ei) observed (O - E) Ei d 45 45.150 -0.150 0.000 32 26.400 5.600 1.188 18 18.750 -0.750 0.030 12 14.550 -2.550 0.447 9 11.850 -2.850 0.685 3 10.050 -7.050 4.946 13 8.700 4.300 2.125 9 7.650 1.350 0.238 9 6.900 2.100 0.639 150 150.000 0.000 10.299 chi-squa 10.30 re df 8 p-value 0.2447 Now, we have = 10.30 is the obtained test statistic value. Here, k is the number of different categories of outcomes = 9 Then degrees of freedom = k - 1 = 9 - 1 = 8 and P-value = 0.2447 (The smaller P-value is, the stronger the evidence against H a0 in favor of H . 1 P-value is small like 0.01 or smaller, we may conclude that the null hypothesis H is 0 strongly rejected in favor of H1 If P-value is between 0.05 P-value 0.01, we may conclude that the null hypothesis H i0rejected in favor of H .1n other cases, i.e., P-value > 0.05, we may conclude that the null hypothesis H is accepted) 0 Since P-value is greater than 0.05 we accept the null hypothesis at 5% level of significance and conclude that these leading digits are from a population of leading digits that conform to Benford’s law.

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