Question

A random sample of 700 voters in a particular city found 266 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

Solution :

Given that,

n = 700

x = 266

Point estimate = sample proportion = = x / n = 266/700=0.38

1 - = 1-0.38=0.62

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96} ( Using z table )   

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.38*0.62) /700 )

E = 0.0360

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.38 -0.0360 < p <0.38+ 0.0360

0.3440< p <0.4160

The 95% confidence interval for the population proportion p is : 0.34,0.42