Castaneda v. Partida is an important court case in which statistical methods wer

Question

Castaneda v. Partida is an important court case in which statistical methods were used as part of a legal argument. When reviewing this case, the Supreme Court used the phrase "two or three standard deviations" as a criterion for statistical significance. This Supreme Court review has served as the basis for many subsequent applications of statistical methods in legal settings. (The two or three standard deviations referred to by the Court are values of the z statistic and correspond to P-values of approximately 0.05 and 0.0026.) In Castaneda the plaintiffs alleged that the method for selecting juries in a county in Texas was biased against Mexican Americans. For the period of time at issue, there were 181,025 persons eligible for jury duty, of whom 142,475 were Mexican Americans. Of the 867 people selected for jury duty, 333 were Mexican Americans.

(a) What proportion of eligible voters were Mexican Americans? Let this value be po. (Round your answer to four decimal places.)

b) Let p be the probability that a randomly selected juror is a Mexican American. The null hypothesis to be tested is Ho: p = po. Find the value of p for this problem, compute the z statistic, and find the P-value. What do you conclude? (A finding of statistical significance in this circumstance does not constitute a proof of discrimination. It can be used, however, to establish a prima facie case. The burden of proof then shifts to the defense.) (Use = 0.01. Round your test statistic to two decimal places

z=

c)We can reformulate this exercise as a two-sample problem. Here we wish to compare the proportion of Mexican Americans among those selected as jurors with the proportion of Mexican Americans among those not selected as jurors. Let p1 be the probability that a randomly selected juror is a Mexican American, and let p2 be the probability that a randomly selected nonjuror is a Mexican American. Find the z statistic and its P-value. (Use = 0.01. Round your test statistic to two decimal places

z=

Explanation / Answer

a.
For the period of time at issue, there were 181,025 persons eligible for jury duty,
of whom 142,475 were Mexican Americans

po=142,47/181,025=0.07870

b.
Given that,
possibile chances (x)=333
sample size(n)=867
success rate ( p )= x/n = 0.3841
success probability,( po )=0.384
failure probability,( qo) = 0.616
null, Ho:p=0.384  
alternate, H1: p!=0.384
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.38408-0.384/(sqrt(0.236544)/867)
zo =0.005

c.
Given that,
sample one, x1 =333, n1 =867, p1= x1/n1=0.384
sample two, x2 =534, n2 =867, p2= x2/n2=0.616
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.384-0.616)/sqrt((0.5*0.5(1/867+1/867))
zo =-9.654
| zo | =9.654
critical value
the value of |z | at los 0.01% is 2.576
we got |zo| =9.654 & | z | =2.576
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -9.6539 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
test statistic: -9.654

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