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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

ad=40 x 1350 = 54000

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

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