In: Statistics and Probability
Benford's Law states that the first nonzero digits of numbers drawn at random from a large complex data file have the following probability distribution.†
First Nonzero Digit | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Probability | 0.301 | 0.176 | 0.125 | 0.097 | 0.079 | 0.067 | 0.058 | 0.051 | 0.046 |
Suppose that n = 275 numerical entries were drawn at random from a large accounting file of a major corporation. The first nonzero digits were recorded for the sample.
First Nonzero Digit | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Sample Frequency | 87 | 46 | 34 | 23 | 21 | 18 | 13 | 17 | 16 |
Use a 1% level of significance to test the claim that the distribution of first nonzero digits in this accounting file follows Benford's Law.
A. What is the level of significance?
B. State the null and alternate hypotheses.
C. Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)
D. Are all the expected frequencies greater than 5?
E. What sampling distribution will you use?
F. What are the degrees of freedom?
G. Estimate the P-value of the sample test statistic.
H. Based on your answers in parts (a) to (g), will you reject or fail to reject the null hypothesis of independence?
using minitab>stat>tables >chi square
we have
Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: frequency
Using category names in Data
Test Contribution
Cate Observed Proportion Expected to Chi-Sq
1 87 0.301 82.775 0.215652
2 46 0.176 48.400 0.119008
3 34 0.125 34.375 0.004091
4 23 0.097 26.675 0.506303
5 21 0.079 21.725 0.024194
6 18 0.067 18.425 0.009803
7 13 0.058 15.950 0.545611
8 17 0.051 14.025 0.631061
9 16 0.046 12.650 0.887154
N DF Chi-Sq P-Value
275 8 2.94288 0.938
A. the level of significance is 0.01
B) the null and alternate hypotheses are
Ho:the distribution of first nonzero digits in this accounting file follows Benford's Law
ha:the distribution of first nonzero digits in this accounting file do not follows Benford's Law
C.the value of the chi-square statistic for the sample is 2.94288
D. yes,all the expected frequencies are greater than 5
E. we will use normal sampling distribution
F. the degrees of freedom is 9-1 = 8
G. the P-value of the sample test statistic is 0.938
H. since p value is greater than 0.01 so we are fail to reject the null hypothesis of independence .