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In a survey of 1002 people, 701 said that they voted in a recent presidential election...

In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.

(a) Find a 99% confidence interval estimate of the proportion of people who say that they voted.

(b) Are the survey results consistent with the actual voter turnout of 61%? Why or why not?

Math   Statistics-And-Probability   
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Answer #1

a) = 701/1002 = 0.6996

At 99% confidence interval the critical value is z* = 2.575

The 99% confidence interval for population proportion is

+/- z* * sqrt((1 - )/n)

= 0.6996 +/- 2.575 * sqrt(0.6996 * (1 - 0.6996)/1002)

= 0.6996 +/- 0.0373

= 0.6623, 0.7369

b) since 0.61 doesn't lie in confidence the interval, so the results are not consistent with the actual voter turnout of 61%.

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