# Law, Economics & Statistics

Law,Economics &amp Statistics

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NursesExam

 Passing exams Failing exams Row total Whites 26 0 26 Blacks 4 5 9 Column total 30 5 35

Pearson’sChi–‐squaredtest is used to test the null hypothesis that ethnicity and theexamination pass rate is independent. The formulae to be used is χ2=(∑i (Oi−Ei) 2)/ Ei, where Oi is the observed frequency in the ithe cell of the table and Ei is the expected frequency in the i thecell of the table. The expected cell frequency can be obtained bymultiplying the row total and the column total then dividing theresults by the number of observations.

Inthis case the expected cell frequency for the whites in relation topassing exams is 26/35 x30 = 22.3 and 26/35x 5= 3.7 for failingexams. For the blacks it will be 9/35×30= 7.7 for passing exams and9/35x 5= 1.3 for failing exams. The results of the test will beχ2=(26-22.3)2/2.3 + (0-3.7)2/3.7 + (4-7.7)2/7.7+ (5-1.3)2/1.3 = 22. Ahigh value of χ2 shows that there is an association between passingexams and race. This then rejects the null hypothesis that race andthe examination pass rate is independent.

UsingFisher Exact test, the probability of the above data will becalculated as follows. The probability ofpassing exams will be 30/35= 0.85 while that of failing is 5/35= 0.15The probability values are high and this shows that there is arelationship between passing exams and race among the blacks and thewhites. The null hypothesis stating that race and the examinationpass rate is independent will be rejected.

Death-Qualifiedjuries

Thereis a statistically significant difference in the voting pattern forboth “beforedeliberation”and“after deliberation.”In thefirst-degree murder, before the deliberation, there were 20 votes andafter the deliberation, there were two votes for the death-qualifiedand there was one vote for excludable in both deliberation ballots.For the second-degree murder, there were 55 and 34 votes for eachballot respectively. This shows that there is no relationship betweenthe two voting patterns.

Themethod used to prove this is maximumentropy probability distributionsmodel. Probability distribution will be determined from the data andcolumn with the highest probability will be considered to be havingthe a proper distribution. The probability for death- qualifiedbefore deliberation is 258/ 514= 0.5 and for the excludable is30/514= 0.06. The total probability before deliberation is 0.56.Onthe other hand the probability for death qualified jurors afterdeliberation is 197/514= 0.38 while for excludable is 29/514= 0.05.The total probability after the deliberation is 0.43. Because theprobabilities of both before and after deliberation are not equalthis means that the deliberation matters.

Deliberationchanges the votes of mock jurors because there are no equal votes forboth before and after deliberation for both the death-qualified andexcludable jurors and the probabilities of the of the twodeliberations are not equal showing that there were unequaldistributions of votes before and after deliberation. If there wasone more deliberation more changes will be observed in the votingpatterns.

PoliceExamination

 Pass Fail Row total Hispanic officers 3 23 26 Other officers 14 50 64 Column total 17 73 90

Usingthe Pearsonchi‐squaredmethodtotestthenullhypothesisthatethnicityandtheexaminationpassrateareindependentwill provide the following results.

Theformulae to be used is χ2= (∑i (Oi−Ei) 2)/ Ei

Theexpected cell frequency for the Hispanic officers in relation topassing examsis 26/90 x17 = 5 and for failing exams is 26/90x 73= 21.For the for the other officers it will be 14/90×17= 3 and for failingexams is 64/90x 73= 52. The results of the test will beχ2=(3-5)2/5 + (23-21)2/21 + (14-3)2/3 + (50-52)2/52 = 41.07. A smallvalue of χ2 shows that there is no association between passing examsand ethnicity. This proves that ethnicity and passing exams areindependent. This will fail reject the null hypothesis that ethnicityand examination pass rate are independent.

UsingFisher Exact test, the probability of the above data will becalculated. The probability of passing exams will be 17/90= 0.0.19while that of failing exams will be is 73/90= 0.8. The probabilityvalues are low for passing exams and high for failing exams and thisshows that there is no relationship between passing exams andethnicity among the Hispanic officers and other officers. The nullhypothesis stating that ethnicity and the examination pass rate isindependent will not be rejected.

Usingthe two sample test Z-test method, the value of Z will be arrived atby subtracting X2 from X1 which are the means of the two samples thenthe hypothesized difference will be subtracted from the total. Theresults found will then be divided by the standard deviation of thetwo populations. The means of the two samples are 17/2= 8.5 and 73/2=36.5. The standard deviation will be 2.7 for Hispanic officers and6.7 for other officers. The value of Z will then be(8.5-36.5)/(2.7+6.7)= -2.98. Therefore, because the results are notequal to zero then ethnicity and passing exams are independent. The95% CI is 3/26x 100= 50% for the Hispanic officers 14/64x 100= 22%.