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# Benford’s Law. According to Benford’s law, a variety of ISBN: 9780321836960 18

## Solution for problem 21BSC Chapter 11.2

Elementary Statistics | 12th Edition

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Problem 21BSC

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 Detecting Fraud? When working for the Brooklyn district attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, 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, 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?

Step-by-Step Solution:

Problem 21BSC Step 1 Given, the frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, 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, the check amounts appear to result from fraud. By using = 0.01 level of significance to test for goodness of fit with Benford’s law. The Hypotheses can be expressed as H : p = 0.301, p = 0.176, p = 0.125, p = 0.097, p = 0.079, p = 0.069, p = 0.058, p = 0 1 2 3 4 5 6 7 8 0.051, p = 9.051, p = 0.010 H1 At least one of the proportion is not equal to its claimed value. The Test Statistic here is 2 (Oi Ei) = Ei Where O = Oierved frequency Ei Expected frequency k = number of different categories of outcome n = the total number observed sample value. Here, the leading digits of amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. The expected frequency can be calculated by E = np Observed Expected Frequency 2 Leading digits frequency (O E ) (O E ) 2 (O i E i i i i i E (O) E = np i 55688. 1 0 235.984 -235.984 235.984 45 15125. 2 15 137.984 -122.984 109.6136 06 3 0 98 -98 9604 98 0.0023 4 76 76.048 -0.048 .0000303 04 173942 5 479 61.936 417.064 2808.421 .4 17022. 6 183 52.528 130.472 324.0737 94 1404.1 7 8 45.472 -37.472 30.87946 51 8 23 39.984 -16.984 288.45 7.2142 63 1300.6 9 0 36.064 -36.064 36.064 12 2 (Oi Ei) = Ei = 3650.251 Step 2 At first we need to install X2GOF in TI-83 calculator The following steps involved in finding the test statistic using TI-83 calculator, 1. Press STAT to enter the data, to select the first item in the list “Edit” press 1 in EDIT 2. In list L1 enter the observed frequencies and in list L2 enter the expected frequencies. 3. Now press PRGM and choose X2GOF and press ENTER 4. Enter L1 for the OBS LIST and Enter L2 for EXP LIST and press ENTER Now, we have = 3650.251 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 We need to find the critical using the following table corresponding to 1% level of significance. Ta Area to the ble right of critical A- value 4 . . . . . . 9 .00 Df .995 9 9 9 1 0 .025 .01 9 7 5 0 0 5 5 5 1 1 1 1 2 2 2 4 . . . . 20. . . 18.47 7 .989 2 6 1 8 16.013 27 0 0 5 3 9 6 3 8 1 6 9 0 7 3 7 7 1 1 1 2 2 3 3 5 . . . . 21. 8 1.344 6 1 7 4 . . 17.535 20.09 95 3 5 0 4 8 3 9 5 6 0 6 0 3 0 2 7

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##### ISBN: 9780321836960

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