Respuesta :
Answer:
The null and alternative hypothesis are:
[tex]H_0: \pi=0.804\\\\H_a:\pi<0.804[/tex]
Statistic z=-33
P-value=0
Conclusion: The P-value is much smaller than the significance level. This sample result is so unprobable that allows us to think that the population proportion is different from π=0.804. This sample proportion can not be product of pure chance.
The null hypothesis is rejected.
There is enough evidence to support the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury.
Step-by-step explanation:
We have to perform a test of hypothesis on the proportion.
The claim is that the selection process is biased against allowing this ethnicity to sit on the grand jury. This means that the proportion of people of this ethnic group selected for jury is way below from the proportion that is eligible for jury.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.804\\\\H_a:\pi<0.804[/tex]
The significance level is 0.05.
The sample we will use is the total of 896 people that were selected for grand jury duty (n=896). 37% of them were from the ethnicity for which we are testing, so the sample proportion is p=0.37.
The standard error is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.804*0.196}{896}}=\sqrt{0.0001759}=0.013[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.37-0.804+0.5/896}{0.013}=\dfrac{-0.433}{0.013}=-33.34[/tex]
The P-value of this test statistic is P=0
[tex]P-value=P(z<-33)=0[/tex]
The P-value is much smaller than the significance level. This sample result is so unprobable that allows us to think that the population proportion is different from π=0.804. This sample proportion can not be product of pure chance.
The null hypothesis is rejected.
There is enough evidence to support the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury.
