A polling company produces a poll with a 5% margin of error. How many additional people must they survey to reduce the margin of error to no more than 3%?
We assume that the poll is looking at a proportion (e.g. the percentage of people with advanced degrees or percent of people supporting a bill, etc...)
For sufficiently large samples, we can assume a normal distribution.
Since we are not given the population variance, we let `hat(p)=.5 ` since this produces the largest variance and will require the largest sample.
The sample size `n ` for a proportion can be found by:
`n=hat(p)hat(q)(z_(alpha/2)/E)^2 ` where `hat(q)=1-hat(p) ` , ` E ` is the error, and `z_(alpha/2) ` is the confidence level.
We are given `E=.05 ` ; assuming we use the typical `alpha=.05 ` or 95% confidence we can solve for n:
`n=(.5)(.5)(1.96/.05)^2 =384.16` so the original sample was 385 people.
For a 3% error we can solve for n:
`n=(.5)(.5)(1.96/.03)^2=1067.11 ` so we need a sample of 1068.
The difference in the sample sizes is 683 so we need to interview an additional 683 people.
Note that if we make no assumptions about the normality of the population we can use the fact that for sufficiently large samples, the sample means will be approximately normal. Using the Wald method we get a 95% confidence interval with a sample of 100 for E=.1, about 400 for an E=.05 and about 1000 for an E=.03 for an increase of 600 people.