A Gallup Poll asked a sample of Canadian adults if they thought the law should a
ID: 3149590 • Letter: A
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 284 people in Quebec, 216 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^^^.
Show work and explain please.
Explanation / Answer
Given that 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 284 people in Quebec, 216 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?
c = confidence level = 95% = 0.95
sample proportion (p) = 216 / 284 = 0.761
q = 1 - p = 1 - 0.761 = 0.239
margin of error = z*sqrt [ (p*q)/n ]
z we can find by using EXCEL.
=NORMSINV(probability)
where probability = 1-a/2
a = 1 -c
z = 1.96
margin of error = 1.96*sqrt[ (0.761*0.239) / 284 ]
Margin of error = 1.96*sqrt(0.000641) = 1.96*0.02532 = 0.049
b) Suppose the researchers wanted a margin of error of 0.01 instead. How large a sample would be needed to achieve this?
Margin of error = 0.01
Here we have to find n.
n = p*q (Z/E)2
= 0.761*0.239 *(1.96/0.01)2
= 0.182*38416
= 6996
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