A proposal to amalgamate the two towns of Palookatown and Smallville into one mu

Question

A proposal to amalgamate the two towns of Palookatown and Smallville into one municipality is scheduled to be put to a referendum vote at die next local election. A random survey of 100 voters in each town is conducted, with 57 voters in Palookatown indicating their support for the proposal, and 43 voters in Smallville indicating their support. a) Calculate 90% and 95% confidence intervals for the difference between the levels of support for amalgamation between the two towns. b) Comment oil whether or not the results from Part (a) support the idea that one town is more supportive, overall, of the amalgamation proposal. c) Redo Part (a) at LOC = 99%. d) Redo Part (b) based on the answer from Part (c).

Explanation / Answer

n1=100, p1=0.57=proportion of supprot of palookatown

n2=100, p2=0.43=proportion of support of smallville

Var(p1)=p1(1-p1)/n1=0.57*(1-0.57)/100=0.002451

var(p2)=p2(1-p2)/n2=0.43*(1-0.43)/100=0.002451

(p1-p2)=(0.57-0.43)=0.14

var(p1-p2)=var(p1)+var(p2)=0.002451+0.002451=0.004902

SE(p1-p2)=sqrt(var(p1-p2))=sqrt(0.004902)=0.07

(1-alpha)*100% confidence interval for difference of proportion=(p1-p2)± z(alpha/2)*SE(p1-p2)

(a)90% confidence interval f=0.14±z(0.1/2)*0.07=0.14±1.6449*0.07=0.14±0.1151=(0.0249, 0.2551)

95% confidence interval =0.14±z(0.05/2)*0.07=0.14±1.96*0.07=0.14±0.1372=(0.0028,0.1772)

(b) here we use z-test and z=(p1-p2)/SE(p1-p2)=0.14/0.07=2

critical z(0.05)=1.96 is less than calculate z=2, so one town is more supportive at 90% confidence(or at alpha=0.1) and 95% confidence(or at alpha=0.05)

(c)99% confidence interval =0.14±z(0.01/2)*0.07=0.14±2.5758*0.07=0.14±0.1802=(-0.0403,0.3203)

(d) here critical z(0.01)=2.5758 is more than calculated z=2, so one town is not more supportive at 99% confidence or alpha=0.01

alpha is level of significance

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