Benford's law states that the probability distribution of the first digits of many items (e.g. populations and expenses) is not uniform, but has the probabilities shown in this table.Business expenses tend to follow Benford's Law, because there are generally more small expenses than large expenses.Perform a "Goodness of Fit" Chi-Squared hypothesis test (αα = 0.05) to see if these values are consistent with Benford's Law.If they are not consistent, it there might be embezzelment.Complete this table.XObservedFrequency(Counts)Benford'sLaw P(X)ExpectedFrequency(Counts)111.301 214.176 321.125 49.097 511.079 65.067 76.058 87.051 96.046 Conclusion:O There is sufficient evidence to warrant rejection of the claim that these values are consistent withBenford's Law.O There is not sufficient evidence to warrant rejection of the claim that these values are consistentwith Benford's Law.O The sample data supports the claim that these values are consistent with Benford's Law.There is not sufficient data to support the claim that these values are consistent with Benford's Law.Enter the critical value, along with the significance level and degrees of freedom x (a,df) below thegraph. (Graph is for illustration only. No need to shade.)X²- Distribution0.120.10.080.060.040.02五红XX2,ProbabilityLE432960T115127133 The sum of the observed frequencies is 90.Ho: P1 = .301, P2 = .176, P3 = .125, P4 = .097, p5 = .079, P6 = .067, P7 = .058, ps = .051, Pg = .046H1: at least one is differentOriginal Claim:O HoOH:Test Statistic =(Round to three decimal places.) P-value =(Round to four decimal places.) Decision:O reject the null hypothesisOfail to reject the null hypothesisO Accept the null hypothesis

Name: Benford's law states that the probability distribution of the first di
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: Benford's law states that the probability distribution of the first digits of many items (e.g. populations and expenses) is not uniform, but has the probabilities shown in this table.Business expenses tend to follow Benford's Law, because there are g

Given significance level=αα =0.05Hypothesis taken in consideration H0: p1=0.301 , p2=0.176,p3=0.125,p4=0.097,p5=0.079,p6=0.067,p7=0.058,p8=0.051,p9=0.046H1: atleast one is different Expected observation = total observed x probability x Observed frequency (counts) Benford's Law P(X) Expected Frequency (counts) (observed frequency-expected frequency)2expected frequency 1 11 0.301 27.09 9.556593 2 14 0.176 15.84 0.213737 3 21 0.125 11.25 8.45 4 9 0.097 8.73 0.008351 5 11 0.079 7.11 2.128284 6 5 0.067 6.03 0.175937 7 6 0.058 5.22 0.116552 8 7 0.051 4.59 1.265381 9 6 0.046 4.14 0.835652 total 90 90 22.75049 Now, Test statistic=Total (obsserved frequency -expectedfrequency)2 expected frequency=22.751 p value = 0.0019 round to 4 decimal places The accepted region according to chi square distribution and with 8 degrees of freedom is 65.79336 and 116.9891 and since our test statistic does not lie in this region therefore we will reject the null hypothesis which means that there is at least one probability is different. Therefore there is sufficient data to warrant rejection of the claim that these values are consistent with Benford's Law