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%?

a)n=543

b)n=1509

c)n=13573

Step-by-step explanation:

a)

c=98%,

Margin Error

but

where the confidence level is 1-α=0.98

cross multiply

where p=0.5

input the values

n=0.33

b) E=0.33

confidence level

n=(2.33/0.33)²×0.5(1-0.5)=1504

c)

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

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total

y=x+9

?

step-by-step explanation: