Question

In a recent survey of 1002 people, 701 said that they voted in a recent presidential election. Voting records show that 61% of eligible voters actually did vote. a) Find the 95% confidence interval estimate of the proportion of people who say that they voted.

Answer 1

The 95% confidence interval estimate of the proportion of people who say that they voted

(0.67122 , 0.72798)

Explanation:

Step(i):-

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

Sample proportion

`p = (x)/(n) = (701)/(1002) = 0.6996`

Step(ii)

The 95% confidence interval estimate of the proportion of people who say that they voted

`(p^(-) - Z_(0.05) \sqrt{(p(1-p))/(n) } , p^(-) + Z_(0.05) \sqrt{(p(1-p))/(n) } )`

`(0.6996 - 1.96\sqrt{(0.6996(1-0.6996))/(1002) } , 0.6996 + 1.96 \sqrt{(0.6996(1-0.6996))/(1002) } )`

(0.6996 - 1.96 X 0.01448 , 0.6996 + 1.96 X 0.01448)

(0.6996 - 0.02838 , 0.6996 + 0.02838)

(0.67122 , 0.72798)

Final answer:-

The 95% confidence interval estimate of the proportion of people who say that they voted

(0.67122 , 0.72798)