Maths6
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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 forgoodnessoffit 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% 

(OE)^{2}/E 
1.011 
0.649 
1.950 
0.007 
3.880 
3.275 
2.703 
0.490 
0.881 
Calculatethe χ^{2} teststatistic.
ChiSquare test Statistic equals {(observed*frequency) –(expected*frequency)}^{2}divided by (expected * frequency)
=1.011 + 0.649+1.95+0.007+3.88+3.275+2.703+0.49+0.881 = 14.846
Calculatethe χ^{2} criticalvalue.
Degreesof freedom = 9 1 = 8
Alpha= 0.05
Thereforethe χ^{2} criticalvalue = 2.73
Isthere sufficient evidence to conclude that the checks are the resultof fraud?
Sinceour x^{2} statistic(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 χ^{2} teststatistic.
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
thereforeχ^{2}={(54000 – 7378)^{2}(1567)}/{257*1384*1567*74}
=3406048255228/41244869104 = 82.58
Calculatethe χ^{2} criticalvalue.
 Degreesof freedom equal (number of columns minus one) x (number of rowsminus one) not counting the totals for rows or columns. For our datathis gives (21) x (21) = 1.
Ourchi square statistic (x^{2} =82.5811),our predetermined alpha level of significance (0.01), and our degreesof freedom (df = 1)
Fromthe chi square table the χ^{2}criticalvalue = 0.0002
Isthere sufficient evidence to reject the defense claim that deaths onshifts are independent of whether Gilbert was working?
Sinceour x^{2} statistic(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). Chi Square Statistics. Retrieved 26 December 2015, fromhttp://math.hws.edu/javamath/ryan/ChiSquare.html