Question

Most polls (performed by market research companies) are given with a margin of error of about 3% (if you read poll results in a national newspaper, you will often see something like "these results are given plus or minus 3 percentage points").

1. If you wish to poll Vancouver residents to determine their level of support for a plan to increase green space in the city, and you wish to use a 98% confidence level, what is the minimum number of residents you should poll to obtain results that have a margin of error of 3%?      
2. Suppose you wanted to obtain the same margin of error, but at a 99% confidence level, how many more residents would you need to sample (how many more than you gave in your answer to question 1)?

Answer 1

Solution,

Given that,

1) $\hat p$ =  1 - $\hat p$ = 0.5

margin of error = E = 0.03

At 98% confidence level

$\alpha$ = 1 - 98%

$\alpha$ = 1 - 0.98 =0.02

$\alpha$/2 = 0.01

Z_$\alpha$/2 = Z_0.01 = 2.326

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (2.326 / 0.03)² * 0.5 * 0.5

= 1502.85

sample size = n = 1503

2) $\hat p$ =  1 - $\hat p$ = 0.5

margin of error = E = 0.03

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (2.576 / 0.03)² * 0.5 * 0.5

= 1843.27

sample size = n = 1844