In the recent survey of 1002 people 701 said that they voted

In the recent survey of 1002 people, 701 said that they voted for in a recent presidential election. Voting records show that 61% of eligible voters actually 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?

Solution

Proportion observed =701/1002 = 0.6996=69.96%

Sample Size: 1002
Observed Proportion: 69.96%
Confidence Level: 99%

z alpha/2 = 2.58

STd error = rt of p(1-p)/n where p = 0.6996 and n = 1002

Margin of error = 2.58 (std error)

= ±3.73

Confidence interval 99% = (66.23% , 73.69%)

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b) Not consistent with actual voter turnout of 61% as 61% is not contained in the 99% confidence interval for sample proporiton.