In a survey of 1000 people, 700 said that they voted in a recent presidential election....

In a survey of 1000 people, 700 said that they voted in a recent presidential election. Voting records show that 61% of eligible voters actually did vote. Construct a 96% confidence interval estimate of the proportion of people who say that voted.

Expert Solution

We have given,

x=700
n=1000
Estimate for sample proportion=0.7
Level of significance is =1-0.96=0.04
Z critical value(using Z table)=2.054

Confidence interval formula is

=(0.6702, 0.7298)

orchestra answered 2 years ago

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. Given that 61% of eligible voters actually did vote, find the probability that among 1002 randomly selected eligible voters, at least 701 actually did vote. What does the result suggest? [Write your answer as a complete sentence]

In a survey of 1359 people, 954 people said they voted in a recent presidential election....

In a survey of 1359 people, 954 people said they voted in a recent presidential election. Voting records show that 68% of eligible voters actually did vote. Given that 68% of eligible voters actually did vote, (a) find the probability that among 1359 randomly selectedvoters, at least 954 actually did vote. (b) What do the results from part (a) suggest? (a) P(X ≥ 954) = ? (Round to four decimal places as needed.) (b) What does the result from part...

In a survey of 1168 people, 741 people said they voted in a recent presidential election....

In a survey of 1168 people, 741 people said they voted in a recent presidential election. Voting records show that 61% of eligible voters actually did vote. Given that 61% of eligible voters actually did vote, (a) find the probability that among 1168 randomly selected voters, at least 741 actually did vote. (b) What do the results from part (a) suggest?

In a survey of 1453 people, 1033 people said they voted in a recent presidential election....

In a survey of 1453 people, 1033 people said they voted in a recent presidential election. Voting records show that 69 % of eligible voters actually did vote. Given that 69 % of eligible voters actually did vote, (a) find the probability that among 1453 randomly selected voters, at least 1033 actually did vote. (b) What do the results from part (a) suggest? (a) P(x ?1033 )= _______ (Round to four decimal places as needed)

In a survey of 1073 people, 743 people said they voted in a recent presidential election....

In a survey of 1073 people, 743 people said they voted in a recent presidential election. Voting records show that 66% of eligible voters actually did vote. Given that 66% of eligible voters actually did vote, (a) find the probability that among 1073 randomly selected voters, at least 743 actually did vote. (b) What do the results from part (a) suggest? (a) P(Xgreater than or equals743)equals nothing (Round to four decimal places as needed.)

In a survey of 1016 people, 778 people said they voted in a recent presidential election....

In a survey of 1016 people, 778 people said they voted in a recent presidential election. Voting records show that 74% of eligible voters actually did vote. Given that 74% of eligible voters actually did vote, (a) find the probability that among 1016 randomly selected voters, at least 778 actually did vote. (b) What do the results from part (a) suggest? (a) P(Xgreater than or equals778)equals nothing (Round to four decimal places as needed.) (b) What does the result...

In a survey of 1301 people, 842 people said they voted in a recent presidential election....

In a survey of 1301 people, 842 people said they voted in a recent presidential election. Voting records show that 62% of eligible voters actually did vote. Given that 62% of eligible voters actually did vote, (a) find the probability that among 1301 randomly selected voters, at least 842 actually did vote. (b) What do the results from part (a) suggest? (a) P(Xgreater than or equals≥842)equals=nothing (Round to four decimal places as needed.) (b) What does the result from part...

In a survey of 1407 people, 1077 people said they voted in a recent presidential election....

In a survey of 1407 people, 1077 people said they voted in a recent presidential election. Voting records show that 74% of eligible voters actually did vote. Given that 74% of eligible voters actually did vote, (a) find the probability that among 1407 randomly selected voters, at least 1077 actually did vote. (b) What do the results from part (a) suggest?

In a survey of 1373 people, 930 people said they voted in a recent presidential election....

In a survey of 1373 people, 930 people said they voted in a recent presidential election. Voting records show that 65% of eligible voters actually did vote. Given that 65% of eligible voters actually did vote, (a) find the probability that among 1373 randomly selected voters, at least 930 actually did vote. (b) What do the results from part (a) suggest? (a) P(Xgreater than or equals930)equals nothing (Round to four decimal places as needed.) (b) What does the result from...

1. In a presidential election, 308 out of 611 voters surveyed said that they voted for...

1. In a presidential election, 308 out of 611 voters surveyed said that they voted for the candidate who won. Use a 0.01 significance level to test that among all voters, the percentage who believe they voted for the winning candidate is different from 43%, which is the actual percentage of votes for the winning candidate. What does the result suggest about voter perceptions? Must show all work.

Question