6. The article \"Statistical Evidence of Discrimination\" (J. of Amer. Stat. Ass

Question

6. The article "Statistical Evidence of Discrimination" (J. of Amer. Stat. Assoc. (1982):773-783) discussed the court case Swain v. Alabama (1965) in which it was alleged that there was discrimination against African Americans in grand jury selection. Census data suggested that 25% of those eligible for grand jury service were African American, yet a random sample of 1050 individuals called to appear for possible duty yielded only 177 African Americans. Using = .01, does this data support a conclusion of discrimination? (In this context it would be considered discrimination if the proportion of African Americans chosen is significantly less than the percentage of eligible people who are African American.)

Explanation / Answer

Set Up Hypothesis
Null, H0:P>0.25
Alternate,African Americans chosen is significantly less than the percentage of eligible H1: P<0.25
Test Statistic
No. Of Success chances Observed (x)=177
Number of objects in a sample provided(n)=1050
No. Of Success Rate ( P )= x/n = 0.1686
Success Probability ( Po )=0.25
Failure Probability ( Qo) = 0.75
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.16857-0.25/(Sqrt(0.1875)/1050)
Zo =-6.0936
| Zo | =6.0936
Critical Value
The Value of |Z | at LOS 0.01% is 2.33
We got |Zo| =6.094 & | Z | =2.33
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Left Tail -Ha : ( P < -6.09356 ) = 0
Hence Value of P0.01 > 0,Here we Reject Ho

we have evidence that proportion of African Americans chosen is significantly
less than the percentage of eligible people who are African American

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