Question

Most polis (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 80% 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 90% 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 80% confidence level

$\alpha$ = 1 - 80%

$\alpha$ = 1 - 0.80 =0.20

$\alpha$/2 = 0.10

Z_$\alpha$/2 = Z_0.10 = 1.282

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

= (1.282 / 0.03)² * 0.5 * 0.5

= 456.53

sample size = n = 457

2) $\hat p$ =  1 - $\hat p$ = 0.5

margin of error = E = 0.03

At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

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

= (1.645 / 0.03)² * 0.5 * 0.5

= 751.67

sample size = n = 752