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 285 people in Quebec, 207 of whom agreed that doctor-assisted suicide should be allowed.

(a) What is the margin of error of the large-sample 99.5% confidence interval for the proportion of all Quebec adults who would allow doctor-assisted suicide?
MoE:

(b) How large a sample is needed to get a ±3±3 percentage point margin of error (this is very commonly used)? Use the previous sample as a pilot study to get p*.
(You may need four decimal places in your critical value to solve this problem.)
Sample size:

Answer 1

Solution :

Given that,

a) n = 285

x = 207

Point estimate = sample proportion = = x / n = 207 / 285 = 0.726

1 - = 1 - 0.726 = 0.274

At 99.5% confidence level

= 1 - 99.5%

=1 - 0.995 =0.005

/2 = 0.0025

Z/2 = Z_{0.0025 = 2.8070}

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

= 2.8070 (((0.726 * 0.274) / 285)

MoE = 0.074

b) margin of error = E = 0.03

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

= (2.8070 / 0.03)² * 0.726 * 0.274

= 1741.52

sample size = n = 1742