Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. You are planning a sample survey to determine what percent of the voters intend to vote for Chavez. You will contact a SRS of registered voters in the city. You want to estimate the proportion p of Chavez voters with 95% confidence and a margin of error no greater than 3%, or 0.03. How large a sample do you need? In the absence of any earlier poll to start from, use the guess p* = 0.5. n =__________

You need to calculate sample size in order to estimate the population proportion of people that voted for Gloria Chavez for mayor using a 95% confidence level, an estimated proportion of 0.5 and a margin error no greater than 0.03.

The general structure of the CI for the population proportion is "point estimator" ± "margin of error"

The formula for the interval is

p' ± [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex]

Where d= [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex] is the margin of error of the interval.

Now you have to clear the sample size, n, from the formula.

d= [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex]

[tex]\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'()1-p'}{n} }[/tex]

[tex](\frac{d}{Z_{1-\alpha /2}} )^2= \frac{p'(1-p')}{n}[/tex]

[tex]n*(\frac{d}{Z_{1-\alpha /2}} )^2= p'(1-p')[/tex]

[tex]n= (p'(1-p'))*(\frac{Z_{0.975}}{d} )^2= (0.5*0.5)*(\frac{1.965}{0.03} )^2= 1072.5625 = 1073[/tex]

To estimate the proportion of people that voted for Gloria Chavez you need to take a sample of n= 1073 voters.

I hope it helps!