Thread: statistics problem need help ASAP please.

In preparing a report on the economy we need to estimate the proportion of businesses that
intend to hire additional employees in the next 60 days.
(a) If we have no prior estimate of the proportion, how many randomly selected employers
must we sample in order to be 98% confident that our estimation error will be no more
than 0.05?

Originally Posted by kristen7

In preparing a report on the economy we need to estimate the proportion of businesses that
intend to hire additional employees in the next 60 days.
(a) If we have no prior estimate of the proportion, how many randomly selected employers
must we sample in order to be 98% confident that our estimation error will be no more
than 0.05?

When you are asked to determine the minimum sample size required to estimate a population proportion (percentage) within a certain margin of error

n >= ((p*q)/d^2) * Z^2

where:
p is the proportion you want to estimate, and the sample proportion is given in the problem
q = 1-p
d = margin of error, usually given as a "+/-" percentage points
Z = z score corresponding to the desired level of confidence

Since you have no prior estimate of p, then use p = 0.5 (then q = 1-p = 0.5), d = .05, z = 2.326