Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 588 randomly selected Americans surveyed, 354 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between ___?__ and __?___

b. If many groups of 588 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About ___?___ percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about ___?___ percent will not contain the true population proportion.

Answer 1

sample success x =		354
sample size          n=		588
sample proportion p̂ =x/n=		0.6020
std error se= √(p*(1-p)/n) =		0.0202
for 90 % CI value of z=		1.645
margin of error E=z*std error   =		0.0332
lower bound=p̂ -E                       =		0.5688
Upper bound=p̂ +E                     =		0.6352
from above 90% confidence interval for population proportion =(0.5688 ,0.6352)

b)

If many groups of 588 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 90%   of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 10% will not contain the true population proportion