Question

part b: what is the margin of error for this sample?

part c: is there any evidence that this proportion has changed since 2008 based on this sample?

According to a health organization in a certain country, 21.8% of the population smoked in 2008. In 2014, a random sample of 750 citizens of this country was selected, 141 of whom smoked. Complete parts a through c below. Nam a. Construct a 99% confidence interval to estimate the actual proportion of people who smoked in this country in 2014 Due: A00% confidence interval to estimate the actual proportion has a lower limit of and an upper limt of Curre (Round to three decimal places as needed)

Answer 1

x                    =				141
sample size		n                    =		750.0
sample proportion     p̂		x/n=		0.188
std error       =Se		=√(p*(1-p)/n) =		0.0143
for 99 % CI value of z=				2.58
margin of error E=z*std error                            =				0.037
lower confidence bound=sample proportion-margin of error				0.151
Upper confidence bound=sample proportion+margin of error				0.225

a) 99% confidence interval .....has a lower limit of 0.151 and and an upper limit of 0.225

b)margin of error =0.037

c) No, as 21.8% falls in this interval