Question

in a recent court case it was found that during a period of 11 years 890 people were selected for grand jury duty and 43% of them were from the same ethnicity. among the people eligible for grand jury duty, 79.7% 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
which of the following is the hypothesis test to be conducted?
a ( h_{0}: p = 0.797 )
( h_{1}: plt0.797 )
b ( h_{0}: p = 0.797 )
( h_{1}: p
eq0.797 )
c ( h_{0}: plt0.797 )
( h_{1}: p = 0.797 )
d ( h_{0}: p = 0.797 )
( h_{1}: pgt0.797 )
e ( h_{0}: pgt0.797 )
( h_{1}: p = 0.797 )
f ( h_{0}: p
eq0.797 )
( h_{1}: p = 0.797 )
what is the test statistic?
( z=-27.19 )
(round to two decimal places as needed)
what is the p - value?
p - value ( =0.000 )
(round to three decimal places as needed)
what is the conclusion on the null hypothesis?
fail to reject the null hypothesis because the p - value is less than or equal to the significance level, ( alpha )
reject the null hypothesis because the p - value is greater than the significance level, ( alpha )
fail to reject the null hypothesis because the p - value is greater than the significance level, ( alpha )
reject the null hypothesis because the p - value is less than or equal to the significance level, ( alpha )

Explanation:

Step1: Hypothesis Test Selection

The claim is that the selection process is biased against allowing the ethnicity to sit (i.e., \(p < 0.797\)). The null hypothesis \(H_0\) is \(p = 0.797\) and the alternative hypothesis \(H_1\) is \(p < 0.797\), so option A is correct.

Step2: Test Statistic

Given \(n = 890\), \(\hat{p}=0.43\), \(p = 0.797\), \(q=1 - p=0.203\).
The formula for the test - statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}\)

$$
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$$

Step3: P - value

For \(z=-27.19\), using the standard normal distribution (since the test is left - tailed), the \(P\) - value is the area to the left of \(z =-27.19\). Looking up in the standard normal table (or using a calculator with normalcdf\((-\infty,-27.19)\)), \(P\) - value \(\approx0.000\)

Step4: Conclusion on Null Hypothesis

Since the \(P\) - value (\(0.000\)) \(\leq\alpha = 0.05\), we reject the null hypothesis.

Answer:

• Hypothesis test: A. \(H_0:p = 0.797\), \(H_1:p<0.797\)
• Test statistic: \(z=-27.19\)
• \(P\) - value: \(0.000\)
• Conclusion on null hypothesis: Reject the null hypothesis because the \(P\) - value is less than or equal to the significance level \(\alpha\)
• Final conclusion: There is sufficient evidence to support the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury.