1. In a survey of 1002 people, 701 said that they voted in a recent presidential
ID: 3364590 • Letter: 1
Question
1. In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group (a) Find a 99% confidence interval estimate of the proportion of people who say that they voted. Write your answer in English. (b) Find the margin of error for the 99% confidence interval. (c) Voting record shows that 61% of eligible voters actually did vote. Base on your answer in part (a), are the survey results consistent with the actual voter turnout? Why or why not?Explanation / Answer
Solution:
a) Given that x = 701, n = 1002
= 0.01
p = x/n = 701/1002 = 0.6996
Standard Error of proportion SE= sqrt[p(1-p)/n]
= sqrt[(0.6996)(0.3004)/1002] = 0.0145
Margin Error E = Z *sqrt[p(1-p)/n]
= 2.575*0.0145
= 0.0373
99% confidence interval:
C.I = p ± Margin of Error
= 0.6996 ± 0.0373
= (0.6623, 0.7369)
b) Margin Error E = Z *sqrt[p(1-p)/n]
= 2.575*0.0145
= 0.0373
c) for interval contains values above 61% ; therefore survey is not consistent with the acutal voter turn out.
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