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 98​% confident with a margin of error of 88​%?

​b) Suppose we want to reduce the margin of error to 55​%. 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​%?

Answer 1

Solution :

Given that,

$\hat p$ = 1 - $\hat p$ = 0.5

a) margin of error = E = 0.08

At 98% confidence level

$\alpha$ = 1 - 98%

$\alpha$ =1 - 0.98 =0.02

$\alpha$/2 = 0.01

Z$\alpha$/2 = 2.326

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

= (2.326 / 0.08)² * 0.5 * 0.5

= 211.33

sample size = n = 212

b) margin of error = E = 0.05

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

= (2.326 / 0.05)² * 0.5 * 0.5

= 541.02

sample size = n = 542

b) margin of error = E = 0.01

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

= (2.326 / 0.01)² * 0.5 * 0.5

= 13525.69

sample size = n = 13526