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

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places