Question

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 file 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.

Click on the datafile logo to reference the data.

(a)	Choose the 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%.
	H0: - Select your answer -p > 0.577p = 0.577p < 0.577p ≠ 0.577Item 1 Ha: - Select your answer -p > 0.577p = 0.577p < 0.577p ≠ 0.577Item 2
(b)	Estimate the proportion of stay-at-home residents in Arkansas. If required, round your answer to four decimal places.
	Does this proportion differ significantly from the mean proportion for all states? Use α = 0.05.
	The input in the box below will not be graded, but may be reviewed and considered by your instructor.
(c)	Estimate the proportion of stay-at-home residents in Virginia. If required, round your answer to four decimal places.
	Does this proportion differ significantly from the mean proportion for all states? Use α = 0.05.
	The input in the box below will not be graded, but may be reviewed and considered by your instructor.
(d)	Would you expect the proportion of stay-at-home residents to be higher in Virginia than in Arkansas?
	- Select your answer -YesNoItem 7
	Support your conclusion with the results obtained in parts (b) and (c).
	The input in the box below will not be graded, but may be reviewed and considered by your instructor.

Answer 1

Solution

(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%.

Answer : Hypothesises are

H₀ : Percentage of stay-at-home residents in the two states is same as the overall average of 57.7%. p = 0.577

H_a : Percentage of stay-at-home residents in the two states is different from the overall average of 57.7%. p ≠ 0.577

(b)  Estimate the proportion of stay-at-home residents in Arkansas. Does this proportion differ significantly from the mean proportion for all states? Use = 0.05.

Answer : Proportion of  stay-at-home residents in Arkansas p = 74/120 = 0.6167

so here standard error of standard proportion se₀ = sqrt [0.577 * 0.423/120] = 0.0451

so Test statistic

Z= (p - p₀ )/ se₀ = (0.6167 - 0.577)/ 0.0451 = 0.88

p - value = 2 * Pr(Z > 0.88) = 2 * [ 1 - Pr(Z < 0.88)] = 2 * (1 - 0.8106) = 2 * 0.1894 = 0.3788 > 0.05

so wee can't reject the null hypothesis and conclude that Percentage of stay-at-home residents in theArakansas is same as the overall average of 57.7%.

(3) Estimate the proportion of stay-at-home residents in Virginia. Does this proportion differ significantly from the mean proportion for all states?

Answer :

Proportion of  stay-at-home residents in Arkansas p = 89/180 = 0.4944

so here standard error of standard proportion se₀ = sqrt [0.577 * 0.423/180] = 0.0368

so Test statistic

Z= (p - p₀ )/ se₀ = (0.4944 - 0.577)/ 0.0368 = -2.245

p - value = 2 * Pr(Z < -2.245) = 2 * 0.0124 = 0.0248< 0.05

so wee can reject the null hypothesis and conclude that Percentage of stay-at-home residents in the Virginia is different as the overall average of 57.7%.

(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).

Answer : No, we dont expect that the proportion  of stay-at-home residents to be higher in Virginia than in Arkansas. It is because in part (b) we cant reject the null hypothesis but in part (c) we will be able to reject the null hypothesis so that we means the stay-at-home residents in Virginia is less than the stay-at-home residents in arakansa.

NOTE : In excel you can solve this with the help of sorting function. Just select the data and sort it.

Aster sorting, you can calculate p - vlaue by function 2 * NORMSINV(Z) put z value in it and get the result.