Question

A random sample of 1300 voters in a particular city found 273 voters who voted yes on proposition 200. Find a 95% confidence interval for the true percent of voters in this city who voted yes on proposition 200. Express your results to the nearest hundredth of a percent.

Answer 1

Answer:

n= 1300 , x= 273

c = 95%

formula for confidence inteval is

where Zc is the z critical value for c= 95%

Zc= 1.96

0.18786 < P < 0.23214

convert to percentage by multipling by 100

18.786 < P < 23.214

round to nearest hundredth

18.79 < P < 23.21

confidence interval for the true percent of voters in this city who voted yes on proposition 200 is = ( 18.79 , 23.21 )