Question

In a recent court case it was found that during a period of 11 years 869 people were selected for grand jury duty and 42​% of them were from the same ethnicity. Among the people eligible for grand jury​ duty, 80.4​% were of this ethnicity. Use a 0.05 significance level to test the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.

Find the Test Statistic:

Find the p-value:

Answer 1

Solution:

Given:

n = 869

P hat = 0.42

Among the people eligible for grand jury​ duty, 80.4 % were of this ethnicity.

p = 0.804

Significance level = 0.05

Claim:  the selection process is biased against allowing this ethnicity to sit on the grand jury.

Part a) Identify the null​ hypothesis, alternative​ hypothesis,

H0: p = 0.804 Vs H1: p < 0.804

Part b) test​ statistic

Z =(0. 42 - 0.804)/(sqroot(0.804(1-0.804)/869)

Z = - 28.516

Part c) P-value

P-value = P( Z< z test statistic)

P-value = P( Z< -28.52)

that is: P(Z < -31.39) =0.0000

thus

P-value = 0.0000

Part d) conclusion about the null​ hypothesis

Decision Rule:
Reject H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0

Since P-value = 0.0000 < 0.05 level of significance, we reject the null​ hypothesis.

Part e) final conclusion that addresses the original claim:

Since we have rejected null hypothesis H0, at 0.05 significance level, there is sufficient evidence to support the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury

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