Question

Question 1

In a poll of 725 voters in a campaign to eliminate non-returnable beverage containers, 510 of the voters were opposed. Develop a 95% confidence interval estimate for the proportion of all the voters who opposed the container control bill.

Answer 1

Solution :

Given that,

n = 725

x = 510

Point estimate = sample proportion = = x / n = 0.703

1 - = 0.297

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.703*0.297) / 725)

= 0.033

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.703 - 0.033 < p < 0.703 + 0.033

0.670 < p < 0.736

The 95% confidence interval estimate for the proportion of all the voters who opposed the container control bill is :

(0.670 , 0.736)