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 8%? b) Suppose we want to reduce the margin of error to 5%. 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%?

(Round up to the nearest whole​ number.)

I need help with A,B and C Thank you

Answer 1

Solution :

Given that,

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

margin of error = E = 0.08

a) 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

c) 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