Question

In a recent court case it was found that during a period of 11 years 853 people were selected for grand jury duty and 39​% of them were from the same ethnicity. Among the people eligible for grand jury​ duty, 79.8% were of this ethnicity. Use a 0.01 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.

Which of the following is the hypothesis test to be​ conducted?

A.

Upper H 0 : p greater than 0.798H0: p>0.798

Upper H 1 : p equals 0.798H1: p=0.798

B.

Upper H 0 : p equals 0.798H0: p=0.798

Upper H 1 : p not equals 0.798H1: p≠0.798

C.

Upper H 0 : p equals 0.798H0: p=0.798

Upper H 1 : p less than 0.798H1: p<0.798

Upper H 0 : p not equals 0.798H0: p≠0.798

Upper H 1 : p equals 0.798H1: p=0.798

E.

Upper H 0 : p equals 0.798

What is the test​ statistic?

Z=

​(Round to two decimal places as​ needed.)

What is the​ P-value?

​P-value=

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ : p = 0.798

H_a : 0.798

n = 853

= 39% = 0.39

P₀ = 0.798

1 - P₀ = 1 - 0.798 = 0.202

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.39 - 0.798 / [(0.798 * 0.202) / 853]

= -29.68

Test statistic = -29.68

This is the two tailed test .

P(z < -29.68) = 0

P-value = 2 * 0 = 0