Question

A Student Union executive wants to estimate the proportion of homeowners who live in a sub-division that neighbors the university are opposed to city council move that removes residential restrictions that would allow for more secondary-suites.

This SU executive hires you to take a poll of nn homeowners in the sub-division. You wish to be 99% confident in your sample. In addition, you want to estimate pp to within 0.031 of the true value of pp.

(a) How large of a random sample should you take? (When values from a distribution, use three decimals.)

(b) You decide to take a small simple random sample of n=7n=7 home-owners, of which 22 indicated they are against more secondary-suites. From this information, how many homeowners should you randomly sample to estimate pp? Use the same level of confidence and error as you did in (a)

Answer 1

a)

here margin of error E =	0.031
for99% CI crtiical Z          =	2.576
estimated proportion=p=	0.500
required sample size n =	p*(1-p)*(z/E)2=	1727.00

( please try 1732 if this comes wrong due to rounding error)

b)

here proportion p=2/7=0.286

required sample size n =

p*(1-p)*(z/E)2=

1410.00

( please try 1414 if this comes wrong due to rounding error)