The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623,0.7369).
Given sample size=1002 and confidence level of 99%.
In a sample with a number n of people surveyed with a probability of a success of π and a confidence level of 1-α, we have the following confidence interval of proportions.
π±z[tex]\sqrt{}[/tex]π(1-π)/n
in which z is the z score that has a p value of 1-α/2
n=1002,π=701/1002
So α=0.01, z is the value of Z that has a p value of 1-0.01/2=0.995, so Z=2.575.
Lower limit=π-[tex]z\sqrt{}[/tex]π(1-π)/n
=0.6996-2.575[tex]\sqrt{0.6996*0.3004/1002}[/tex]
=0.6623
Upper limit=π+[tex]z\sqrt{}[/tex]π(1-π)/n
= 0.6996+2.575[tex]\sqrt{0.6996*0.3004/1002}[/tex]
=0.7369
Hence the 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623,0.7369).
Learn more about z test at https://brainly.com/question/14453510
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