All one problem, thank you In a trial, the defendant was accused of issuing chec
ID: 2907463 • Letter: A
Question
All one problem, thank you In a trial, the defendant was accused of issuing checks to a nonexistent vendor. The amounts of the checks are listed in the accompanying data table in order by row. When testing for goodness-of-fit with the proportions expected with Benford's law, it is necessary to combine categories because not all expected values are at least 5. Use one category with leading digits of 1, a second category with leading digits of 2,3, 4, 5, and a third category with leading digits of 6, 7, 8,9. Usinga 0.05 significance level, is there sufficient evidence to conclude that the leading digits on the checks do not conform to Benford's law? 1. Click the icon to view information and the distribution table for Benford's law 2 Click the icon to view the data table of the check amounts. Click here to view the chi-square distribution table.3 Let p1 represent the proportion of checks that have a leading digit of 1. Let p2 represent the proportion of checks that have a leading digit of 2, 3, 4, or 5. Let p3 represent the proportion of checks that have a leading digit of 6, 7, 8, or 9. Assume that all the conditions required for a goodness-of-fit test are satisfied. Identify the null and alternative hypotheses. OA. Ho: P1-0.301 and p2 0.477 and p3 0.222 H1: P1 0.301 and p2 #0.477 and p3 0.222 Ho: p1-0.301 and p2 0.176 and p3 0.125 H1: At least one of the proportions is not equal to the given claimed value. B. c. Ho: P1 0.301 and p2 0.477 and p3 -0.222 Hy: At least one of the proportions is not equal to the given claimed value. Calculate the test statistic (Round to three decimal places as needed.) Find the critical value. Round to three decimal places as needed.)Explanation / Answer
p1 = leading with 1
p2 = leading with 2,3,4 or 5
p3 = leading with 6-9
hypothesis
option C) is correct {note that probability under null is calculated using benford's law}
O1 = 2
O2 = 12
O3 = 9
TS = 6.5669 = 6.567
critical value = 5.991464 = 5.991
since TS > critical value
we reject the null hypothesis
there is sufficient evidence
it appears
Oi Ei (Oi-Ei)^2/Ei 0.301 2 6.923 3.500784198 0.477 12 10.971 0.096512715 0.222 9 5.106 2.969689777 1 23 TS 6.56698669 critical 5.991464547