Question

A biotechnology firm is planning its investment strategy for future products and research labs. A poll found that 11​% of a random sample of 1047 adults approved of attempts to clone a human.

Question: Find the margin of error for this poll if we want 90​% confidence in our estimate of the percent of adults who approve of cloning humans.

ME = ____________ (round to three decimal places as needed.)

Question: Find the margin of error if we want 99% confidence in our estimate.

ME= ________ ( round to three decimal places as needed.)

Answer 1

Solution :

Given that,

n = 1047

= 0.110

1 - = 1 - 0110 = 0.890

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.110 * 0.890) / 1047)

= 0.016

Margin of error = E = 0.016

b ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.110 * 0.890) /1047)

= 0.025

Margin of error = E = 0.025