Question

Suppose election polling with 1000 respondents in a particular state has 7.0% of the sample saying they will vote for the Green Party Candidate. Assuming that it is a fair and accurate representation of the likely voters, what is the 99% confidence interval for the population proportion of rates in the state likely to vote for the green party?

Answer 1

Solution:

Given:

Sample size = n = 1000

Sample proportion of respondents saying they will vote for the Green Party Candidate = 7.0% = 0.07

That is:

We have to find the 99% confidence interval for the population proportion of rates in the state likely to vote for the green party.

Formula:

where

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus

or

_{(Round final answer to specified number of decimal places)}

Thus we are 99% confident that the population proportion of rates in the state likely to vote for the green party is between 4.92% and 9.08%.