Question

How many randomly selected employers must we contact in order to create an estimate in which we are 95​% 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​%?

Answer 1

Answer:

a)n=543

b)n=1509

c)n=13573

Step-by-step explanation:

a)

c=98%,

$E=0.05$

Margin Error $E=Zα/2√p(1-p)/n$

but $n=((Zα/2)/n)²×p(1-p)$

where the confidence level is 1-α=0.98

cross multiply

$Zα/2=2.33$

where p=0.5

input the values

$n=(2.33/0.55)²×0.5(1-0.5)=543$

n=0.33

b) E=0.33

$E=Zα/2√p(1-p)/n$

$n=((Zα/2)/n)²×p(1-p)$

$1-α=0.01$ confidence level

n=(2.33/0.33)²×0.5(1-0.5)=1504

$n=1504$

c) $E=Zα/2√p(1-p)/n$

$n=((Zα/2)/n)²×p(1-p)$

1-α=0.98

cross multiply

Zα/2=2.33

p=0.5

n=(2.33/0.01)²×0.5(1-0.5)=13573

$n=13573$