Question

In preparing a report on the​ economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.

​a) How many randomly selected employers must we contact in order to create an estimate in which we are 99​% confident with a margin of error of 99​%?

​b) Suppose we want to reduce the margin of error to 44​%. What sample size will​ suffice?

​c) Why might it not be worth the effort to try to get an interval with a margin of error of 11​%?

In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. a) How many randomly selected employers must we contact in order to create an estimate in which we are 99% confident with a margin of error of 9%? b) Suppose we want to reduce the margin of error to 4%. What sample size will suffice? c) Why might it not be worth the effort to try to get an interval with a margin of error of 1%? 囲Click on the icon to view a table of the critical values for common confidence levels.

Answer 1

a) p = q = 0.50

For 99% confidence, z = 2.576

E = 0.09

Hence,

Sample size

n = 205

b) p = q = 0.50

For 99% confidence, z = 2.576

E = 0.04

Hence,

Sample size

$n = (0.5)(0.5)(\frac{2.576}{0.04})^2$

n = 1037

c) Because for 1% margin of error, the sample size will be very high i.e. 16590