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 180,975 persons eligible for jury duty, of whom 142,775 were Mexican Americans. Of the 885 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 and your P-value to four decimal places.)

z=

P-value=

(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.)

z=

P-value=

Explanation / Answer

Solution:-

(a) The proportion of eligible voters were Mexican Americans is 0.789.

x = 142,775

n = 180,975

p0 = 142775/180975

p0 = 0.789

b)

p = 333/885

p = 0.3763

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.789

Alternative hypothesis: P 0.789

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.01372

z = (p - P) /

z = - 30.00

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 30.0 or greater than 30.0.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

c)

p1 = 0.3763

p2 = 0.6237

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.50

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.02377

z = (p1 - p2) / SE

z = - 10.41

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 10.41or greater than 10.41.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

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