Question

In a past presidential​ election, 39,885,048. people voted for Candidate​ A; 39,660,675 Candidate​ B; and 193,670...

In a past presidential​ election, 39,885,048. people voted for Candidate​ A; 39,660,675 Candidate​ B; and 193,670 for​ third-party candidates.

a. What percentage of voters chose Candidate​ A? Find the % of voters who chose Candidate A Answer _____%

b. Would it be appropriate to find a confidence interval of voters choosing Candidate​ A? (Answer is A,B,C or D)

A.Yes, it is appropriate to find a confidence interval because the proportion is a sample proportion and the conditions for the Central Limit Theorem are met.

B.​No, it is not appropriate to find a confidence interval because there are three possible outcomes for each observation.

C.​No, it is not appropriate to find a confidence interval because the proportion is a population proportion.

D.No, it is not appropriate to find a confidence interval because the conditions for the Central Limit Theorem are met

0 0
Add a comment Improve this question Transcribed image text
Answer #1

a) Percentage of voters chose Candidate​ A = 39885048 / (39885048 + 39660675 + 193670) = 0.5002 = 50.02%

b) No, it is not appropriate to find a confidence interval because the proportion is a population proportion. Option C is correct.

Add a comment
Know the answer?
Add Answer to:
In a past presidential​ election, 39,885,048. people voted for Candidate​ A; 39,660,675 Candidate​ B; and 193,670...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 5) In a recent presidential election, 500 voters were surveyed and 350 of them said that...

    5) In a recent presidential election, 500 voters were surveyed and 350 of them said that they voted for the candidate who won. a. Construct a 96% confidence interval estimate of the percentage of voters who said they voted for the candidate who won. b. How many voters must they survey if they want 90% confidence that the sample proportion is in error by no more than 0.02?

  • In a survey of 1425 ​people, 1045 people said they voted in a recent presidential election....

    In a survey of 1425 ​people, 1045 people said they voted in a recent presidential election. Voting records show that 71​% of eligible voters actually did vote. Given that 71​% of eligible voters actually did​ vote, (a) find the probability that among 1425 randomly selected​ voters, at least 1045 actually did vote.​ (b) What do the results from part​ (a) suggest?

  • In a survey of 13181318 ​people, 877877 people said they voted in a recent presidential election....

    In a survey of 13181318 ​people, 877877 people said they voted in a recent presidential election. Voting records show that 6464​% of eligible voters actually did vote. Given that 6464​% of eligible voters actually did​ vote, (a) find the probability that among 13181318 randomly selected​ voters, at least 877877 actually did vote.​ (b) What do the results from part​ (a) suggest?

  • In a survey of 1145 ​people, 862 people said they voted in a recent presidential election....

    In a survey of 1145 ​people, 862 people said they voted in a recent presidential election. Voting records show that 73​% of eligible voters actually did vote. Given that 73​% of eligible voters actually did​ vote, (a) find the probability that among 1145 randomly selected​ voters, at least 862 actually did vote.​ (b) What do the results from part​ (a) suggest? ​(a) P(X>862)=

  • In a survey of 1127 ​people, 739 people said they voted in a recent presidential election....

    In a survey of 1127 ​people, 739 people said they voted in a recent presidential election. Voting records show that ​63% of eligible voters actually did vote. Given that ​63% of eligible voters actually did​ vote, (a) find the probability that among 1127 randomly selected​ voters, at least 739 actually did vote.​ (b) What do the results from part​ (a) suggest? (a) P(x≥739) = b) What does the result from part​ (a) suggest? A.People are being honest because the probability...

  • 1107 ​people, 816 people said they voted in a recent presidential election. Voting records show that...

    1107 ​people, 816 people said they voted in a recent presidential election. Voting records show that 71​% of eligible voters actually did vote. Given that 71​% of eligible voters actually did​ vote, (a) find the probability that among 1107 randomly selected​ voters, at least 816 actually did vote.​ (b) What do the results from part​ (a) suggest? ​(a) ​P(X ≥816) ? round 4 decimals

  • In a survey of 1466 people, 1049 people said they voted in a recent presidential election....

    In a survey of 1466 people, 1049 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 1466 randomly selected voters, at least 1049 actually did vote. (b) What do the results from part (a) suggest? (a) P(X2 1049) = (Round to four decimal places as needed.) (b) What does the result from part (a)...

  • In a past presidential election, the actual voter turnout was 62%. In a survey, 882 subjects...

    In a past presidential election, the actual voter turnout was 62%. In a survey, 882 subjects were asked if they voted in the presidential election. Find the mean and standard deviation for the numbers of actual voters in groups of 882. (Round answer to one decimal place.) = 546.84 (Round answer to two decimal places.) 14.42 = Give the interval of usual values for the number of voters in groups of 882. (Enter answer as an interval using square-brackets only...

  • Q 8.19) A candidate is considering nominating for an election to become Mayor of a country...

    Q 8.19) A candidate is considering nominating for an election to become Mayor of a country town. Prior to officially nominating, the candidate decides to conduct a survey to assess the chances of being elected. Ninety voters are randomly selected with 55 indicating they will vote for the candidate. Find a 95% confidence interval for the proportion of voters in the town that will vote for this candidate. Based on this confidence interval, can the candidate determine whether or not...

  • A simple random sample of 300 voters was polled several months before a presidential election. One...

    A simple random sample of 300 voters was polled several months before a presidential election. One of the questions asked was: "Are you satisfied with the choice of candidates for president?" A total of 123 of them said that they were not satisfied. a. What function would you use to create a confident interval for the proportion of voters who are not satisfied with the choice of candidates? b. Create a 99% confident interval for the proportion of voters who...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT