Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

You might think thst if you looked at the first digit in edomly selected mumbers

ID: 3296172 • Letter: Y

Question

You might think thst if you looked at the first digit in edomly selected mumbers that the disribution would be uniform. Actially, ik is not! Simon Newoomb and leter Frank Benfond bod discovered that the digits oocur according to the following distribution: (digit probability) e.White·tatement tothe lawenfotoenest odeiala that will use it to deide whether to pursue the case further or not. Steructure your essay as follows: Given a brief explanation of what a Goodness of Fit test is. Explain why·Goodness of Fit test should be applied in this utation Sme the hypotheses for this situation Interpeet the answer to part c Use the answer to part e to justify the decision in part d Use the decision in part d to make a conclusion about whether the individual is lkely to have mbezaled. Use this to then tell the law enfoecement officials whether thay should pursue the case or not (1,0.301),(2,0.176),(3,0.125),(4,0.097), (5, 0079), (6, 0.067),(7,0.068),(8,0.061), (9,0.04 The IRS eurrontly uscs Benfcet's Law to deteot frasdulent tax data Suppose you work fior the IRS and are investigating an individual suspected of embezzling. The fist digit of 142 checks to a supposed company are as follow Observed Ppints possible: 20 This is attempt 1 of 1 Message instructor about this question Post this question to ferum 3 13 a State the approperiate mall and abemative hypotheses foe this test b.Explain why =0.01 is an appropriate choice forthe level ofsipificance is thi.itsasen. c. What is the P-Value? Report answer to 4 docimal places d. What is your decision? Rejoct the Null Hypohesi Fail to repect en Null I typodsis

Explanation / Answer

Ans:

Chi square test for Goodness of fit:

H0:data fits the Benford's distribution.

H1:data does not fit Benford's distribution.

Total Calculated chi square score=18.775

degree of freedom=9-1=8

p-value=CHIDIST(18.775,8)=0.0161

Significance level=0.01

As, p-value=0.0161>0.01,we fail to reject the Null hypothesis.

Observed(O) Expected(E) (O-E)^2/E pi pi*142 1 35 0.301 42.74 1.402 2 27 0.176 24.99 0.161 3 13 0.125 17.75 1.271 4 16 0.097 13.77 0.360 5 8 0.079 11.22 0.923 6 12 0.067 9.51 0.650 7 9 0.058 8.24 0.071 8 6 0.051 7.24 0.213 9 16 0.046 6.53 13.724 Total 142 1 142 18.775
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote