Question

A Gallup Poll asked a sample of Canadian adults if they thought the law should allow doctors to end the life of a patient who is in great pain and near death if the patient makes a request in writing. The poll included 255 people in Quebec, 203 of whom agreed that doctor-assisted suicide should be allowed.

(a) What is the margin of error for a 95% confidence interval for the proportion of all Quebec adults who would allow doctor-assisted suicide? (Use 3 decimal places)

(b) Suppose the researchers wanted a margin of error of 0.01 instead. How large a sample would be needed to achieve this? Use the previous sample as a pilot study to get a value for p^.

Answer 1

Solution :

n = 255

x = 225

= x / n = 203 / 255 = 0.796

1 - = 1 - 0.796 = 0.204

a ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.796 * 0.204) / 255 )

= 0.049

Margin of error = E = 0.049

b ) Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.01

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.960 / 0.01 )² * 0.5 * 0.5

= 9604

n = sample size = 9604