A random sample of 300 individuals working in a large city indicated that 84 are dissatisfied with their working conditions. Based upon this, compute a 99% confidence interval for the proportion of all individuals in this city who are dissatisfied with their working conditions. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.

What is the lower limit of the 99% confidence interval?

Given a=0.01, |Z(0.005)|=2.58 (from standard normal table)

p=84/300=0.28

So the lower limit of the 99% CI is

p - Z*sqrt(p*(1-p)/n)

=0.28- 2.58*sqrt(0.28*(1-0.28)/300)

=0.21

what is the upper limit of the 99% confidence interval?

p + Z*sqrt(p*(1-p)/n)

=0.28+ 2.58*sqrt(0.28*(1-0.28)/300)

=0.35