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Question1.

Aninvestigator analyzed the leading digits of the amounts from 200checks issued by three suspect companies. The frequencies were foundto be 68, 40, 18, 19, 8, 20, 6, 9, 12 and those digits correspond tothe leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. Ifthe observed frequencies are substantially different from thefrequencies expected with Benford`s law, the check amounts appear tobe the result of fraud. Use a 0.05 significance level to test forgoodness-of-fit with Benford`s law.

FromBenford’s law the probabilities are as follows in respect to theleading digits

 30.1% 17.6% 12.5% 9.7% 7.9% 6.7% 5.8% 5.1% 4.6% Digit 1 2 3 4 5 6 7 8 9 Observed 68 40 18 19 8 20 6 9 12 Expected 60 35 25 19 16 13 12 10 9 Benford’s law 30.1% 17.6% 12.5% 9.7% 7.9% 6.7% 5.8% 5.1% 4.6% (O-E)2/E 1.011 0.649 1.950 0.007 3.880 3.275 2.703 0.490 0.881

Calculatethe χ2&nbspteststatistic.

ChiSquare test Statistic equals {(observed*frequency) –(expected*frequency)}2divided by (expected * frequency)

=1.011 + 0.649+1.95+0.007+3.88+3.275+2.703+0.49+0.881 = 14.846

Calculatethe χ2&nbspcriticalvalue.

Degreesof freedom = 9 -1 = 8

Alpha= 0.05

Thereforethe χ2&nbspcriticalvalue = 2.73

Isthere sufficient evidence to conclude that the checks are the resultof fraud?

Sinceour x2&nbspstatistic(14.846)exceeded the critical value for 0.05 probability level (2.73) weconclude that there is no sufficient evidence to conclude that thechecks are the result of fraud.

Question2

Alertnurses at the Veteran`s Affairs Medical Center in Northampton,Massachusetts, noticed an unusually high number of deaths at timeswhen another nurse, Kristen Gilbert, was working. Kristen Gilbert wasarrested and charged with four counts of murder and two counts ofattempted murder. When seeking a grand jury indictment, prosecutorsprovided a key piece of evidence consisting of the table below.Use a 0.01 significance level to test the defense claim that deathson shifts are independent of whether Gilbert was working.

 Shifts With a Death Shifts Without a Death Gilbert Was Working 40 217 Gilbert Was Not Working 34 1350

Calculatethe χ2&nbspteststatistic.

Contingencytable

 Shifts with death Shifts without death Totals Gilbert Was Working 40 217 257 Gilbert Was Not Working 34 1350 1384 Total 74 1567 1641

bc= 34 * 217 = 7378

thereforeχ2={(54000 – 7378)2(1567)}/{257*1384*1567*74}

=3406048255228/41244869104 = 82.58

Calculatethe χ2&nbspcriticalvalue.

&nbspDegreesof freedom equal (number of columns minus one) x (number of rowsminus one) not counting the totals for rows or columns. For our datathis gives (2-1) x (2-1) = 1.

Ourchi square statistic (x2&nbsp=82.5811),our predetermined alpha level of significance (0.01), and our degreesof freedom (df&nbsp=&nbsp1)

Fromthe chi square table the χ2criticalvalue = 0.0002

Isthere sufficient evidence to reject the defense claim that deaths onshifts are independent of whether Gilbert was working?

Sinceour x2&nbspstatistic(82.58)exceeded the critical value for 0.01 probability level (0.0002) wereject the null hypothesis that the deaths on shifts are independentof whether Gilbert was working

Reference

Math.hws.edu,.(2015).&nbspChi Square Statistics. Retrieved 26 December 2015, fromhttp://math.hws.edu/javamath/ryan/ChiSquare.html