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

Answer:

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

(0.67122 , 0.72798)

Step-by-step explanation:

Step(i):-

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

Sample proportion

                          $p = \frac{x}{n} = \frac{701}{1002} = 0.6996$  

Step(ii)

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

$(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} } )$

$(0.6996 - 1.96\sqrt{\frac{0.6996(1-0.6996)}{1002} } , 0.6996 + 1.96 \sqrt{\frac{0.6996(1-0.6996)}{1002} } )$

(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)