What percentage of the population live in their state of birth? According to the U.S. Census Bureau's American Community Survey, the figure ranges from 25% in Nevada to 78.7% in Louisiana.† The average percentage across all states and the District of Columbia is 57.7%. The data in the DATAfile HomeState are consistent with the findings in the American Community Survey. The data are for a random sample of 120 Arkansas residents and for a random sample of 180 Virginia residents.
(a) Formulate hypotheses that can be used to determine whether the percentage of stay-at-home residents in the two states differs from the overall average of 57.7%.
(a) H0: p ≤ 0.577
Ha: p > 0.577
(b) H0: p > 0.577
Ha: p ≤ 0.577
(c) H0: p ≥ 0.577
Ha: p < 0.577
(d) H0: p = 0.577
Ha: p ≠ 0.577
(e). H0: p < 0.577
Ha: p ≥ 0.577
(B) Estimate the proportion of stay-at-home residents in Arkansas. (Round your answer to four decimal places.)
__________
Does this proportion differ significantly from the mean proportion for all states? Use α = 0.05.
Find the value of the test statistic. (Round your answer to two decimal places.)
––––––––––
Find the p-value. (Round your answer to four decimal places.)
p-value =_________
What is your conclusion?
Reject H0. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.
Reject H0. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.
Do not reject H0. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.
Do not reject H0. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.
(C) Estimate the proportion of stay-at-home residents in Virginia. (Round your answer to four decimal places.)
_________
Does this proportion differ significantly from the mean proportion for all states? Use α = 0.05.
Find the value of the test statistic. (Round your answer to two decimal places.)
_________
Find the p-value. (Round your answer to four decimal places.)p-value =________
What is your conclusion?
Reject H0. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.
Do not reject H0. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.
Do not reject H0. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.
Reject H0. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.
(D)
Would you expect the proportion of stay-at-home residents to be higher in Virginia than in Arkansas? Support your conclusion with the results obtained in parts (b) and (c).
From the results obtained in parts (b) and (c), we would expect the number of stay-at-home residents to be ---Select--- lower higher in Arkansas than Virginia. The sample results show that, rounded to two decimal places, the estimated percentage of Arkansas residents who were born there is_______ %. The sample results also show that, rounded to two decimal places, the estimated percentage of Virginia residents who were born there is______ %.
What percentage of the population live in their state of birth? According to the U.S. Census Bureau's American Community Survey, the figure ranges from 25% in Nevada to 78.7% in Louisiana.† The av...
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