Question

Swain v. Alabama (1965) a lawsuit that argued that there was discrimination against African-Americans in grand jury selection. This case used a statistical argument based on the data that 181 African-Americans out of a random sample of 1075 were called to appear for possible duty.  Using the information that Census data show that 24% of those eligible for jury service were African-America conduct the hypothesis test to test the claim of discrimination. Use a significance level of 0.025.

1. State the null and alternative hypothesis
2. Compute the test statistic and the p-value.
3. Write your conclusion as if you are the expert witness in this case and explain what this p-value means to a jury.
4. What assumptions do we need in order for the conditions of the hypothesis test procedure to be met
5. Calculate a 95% Confidence interval for the true proportion of African-Americans called for grand jury data based on data
6. Use this 95% confidence interval and explain your conclusion as if you are the expert witness in this case and explain what the 95% confidence interval means to the jury.
7. What assumptions do we need in order for the conditions of the 95% confidence interval to be met?

Answer 1

x = 181

n = 1075

= 181/1075 = 0.1684

= 0.24

A.

H₀: p 0.24

H₁: p < 0.24

B.

Formula for test statistic:

z-static: −5.499

p-value: <0.00001

C.

Since p-value = <0.00001 < 0.025 i.e. H₀ is rejected which means the proportion of African-American called to appear for possible duty are less than 24% which is the eligible proportion so there is clear discrimination against African-Americans.

D.

Assumptions:

E.

Confidence interval:

p _Z*

= 0.013

_Z = 1.96

= [0.146,0.191]

F.

If repeated samples are taken of the same kind taken here and confidence intervals are out of them, 95% of those confidence intervals will contain the population proportion.

G.

Same as per D.

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