Question

in a recent court case it was found that during a period of 11 years 876 people were selected for grand jury duty and 35% of them were from the same ethnicity. among the people eligible for grand jury duty, 78.3% 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. ( h_0:p = 0.783 )( h_1:p>0.783 )
b. ( h_0:p = 0.783 )( h_1:p<0.783 )
c. ( h_0:p
eq0.783 )( h_1:p = 0.783 )
d. ( h_0:p = 0.783 )( h_1:p
eq0.783 )
e. ( h_0:p<0.783 )( h_1:p = 0.783 )
f. ( h_0:p>0.783 )( h_1:p = 0.783 )
what is the test statistic?
z=
(round to two decimal places as needed.)

Explanation:

Step1: Determine the hypotheses

The claim is that the selection process is biased against allowing the ethnicity to sit (i.e., the proportion \(p\) of selected from the ethnicity is less than the eligible proportion \(0.783\)).
The null hypothesis \(H_0:p = 0.783\) (no - bias), and the alternative hypothesis \(H_1:p<0.783\) (bias against). So the correct hypothesis test is \(H_0:p = 0.783\), \(H_1:p<0.783\) (Option B).

Step2: Calculate the sample proportion \(\hat{p}\)

Given \(n = 876\) (sample size), and the proportion of selected from the ethnicity \(\hat{p}=0.35\)

Step3: Calculate the test - statistic \(z\)

The formula for the test - statistic \(z\) in a one - sample proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.783\), \(\hat{p}=0.35\), and \(n = 876\)

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Answer:

The hypothesis test to be conducted is \(B. H_0:p = 0.783\), \(H_1:p<0.783\)
The test statistic \(z\approx - 31.15\)