Castaneda v. Partida is an important court case in which statistical methods wer
ID: 3046485 • Letter: C
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 181,725 persons eligible for jury duty, of whom 143,900 were Mexican Americans. Of the 876 people selected for jury duty, 348 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 and your P-value to four decimal places.)
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 = 0.01. Round your test statistic to two decimal places and your P-value to four decimal places.)
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 differentvery similar none of the above
z P-value 18167624Explanation / Answer
Let 181,725 persons eligible for jury duty, of whom 143,900 were Mexican Americans. Of the 876 people selected for jury duty, 348 were Mexican Americans.
Total persons = 181725
Mexican Americans = 143900
selected for jury duty = 876
selected Mexican Americans = 348
A) Proportion of eligible voters were Mexican Americans
p0 = 143900 / 181725
= 0.79
B) probability that a randomly selected juror is a Mexican American (p^)
p^ = 348 / 181725
= 0.0019
Z = 2
p-value = 0.1573
The p-value > alpha (0.05)
So, we Reject the null hypothesis, there is not significant evidence that Mexican Americans are underrepresented on juries.
C) Let p1 be the probability that a randomly selected juror is a Mexican American
p1 = 0.0019
Let p2 be the probability that a randomly selected nonjuror is a Mexican American
Let not selected Mexican American juror = 143900 - 348
= 143552
p2 = 143552 / 181725
= 0.7899
Z = 2
p-value = 0.1573
The p-value > alpha (0.05)
So, Reject the null hypothesis, there is not significant evidence of a difference in proportions
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