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 180,025 persons eligible for jury duty, of whom 143,075 were Mexican Americans. Of the 860 people selected for jury duty, 346 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 p̂ 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 α = 0.01. Round your test statistic to two decimal places and your P-value to four decimal places.)

z

P-value

Conclusion

Reject the null hypothesis, there is significant evidence that Mexican Americans are underrepresented on juries.

Reject the null hypothesis, there is not significant evidence that Mexican Americans are underrepresented on juries.

Fail to reject the null hypothesis, there is not significant evidence that Mexican Americans are underrepresented on juries.

Fail to reject the null hypothesis, there is significant evidence that Mexican Americans are underrepresented on juries.

(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 α = 0.01. Round your test statistic to two decimal places and your P-value to four decimal places.)

z

P-value

Conclusion

Reject the null hypothesis, there is significant evidence of a difference in proportions.

Reject the null hypothesis, there is not significant evidence of a difference in proportions.

Fail to reject the null hypothesis, there is not significant evidence of a difference in proportions.

Fail to reject the null hypothesis, there is significant evidence of a difference in proportions.

How do your answers compare with your results in (b)?

very different

very similar

none of the above

Answer 1

Solution

Part (a)

p₀ = proportion of eligible voters were Mexican Americans

= Mexican Americans eligible for jury duty/total number of persons eligible for jury duty

= 143075/180025

= 0.7948 Answer 1

Part (b)

The value of p_hat

= Mexican Americans selected for jury duty/total number of persons selected for jury duty eligible for jury duty

= 346/860

= 0.4023 Answer 2

Test statistic

Z = (p_hat - p₀)/√{ p₀(1 - p₀)/n}

= (0.4023 – 0.7948)/√(0.7948 x 0.2052/860)

= - 28.50 Answer 3

P-value = P(Z < - 28.50) = 0 Answer 4 [Actual value is: 5.574E-179]

Conclusion

Since p-value less than the given significance level of 0.01, the null hypothesis that p = 0.7948 is rejected against the Alternative: p < 0.7948. Hence, we conclude that

Reject the null hypothesis, there is significant evidence that Mexican Americans are underrepresented on juries. Answer 4

Part (c)

p₁ = the probability that a randomly selected juror is a Mexican American

= 346/860

= 0.4023

p₂ = (143075 – 346)/(180025 – 860)

= 0.7966

Z = (p₁ – p₂)/√[p_hat(1 - p_hat){(1/n₁) + (1/n₂)}] = - 28.5621 Answer 5

where,

n₁ = 860,

n₂ = 179165 and

p_hat = {(n₁ x p_1hat) + (n₂ x p_2hat)}/(n₁ + n₂).

P-value = P(Z < |- 28.56|) = 0 Answer 6 [Actual value is: 1.9895E-179]

Conclusion

Since p-value less than the given significance level of 0.01, the null hypothesis that p₁ = p₁ is rejected against the Alternative: p₁ ≠ p₂. Hence, we conclude that

Reject the null hypothesis, there is significant difference in the proportions Answer 7

Comparison

Conclusions from (b) and (c) are very similar.Answer 8

DONE