Question

A random sample of 78 students were interviewed and 59 said they would vote for a...

A random sample of 78 students were interviewed and 59 said they would vote for a democrat in the 2008 election.

1. Let p represent the proportion of all students at this college who will vote for a democrat. Find a point estimate p for p.

2. Find a 90% confidence interval for p.

3. What assumptions are required for the calculations of part (b)? Do you think these assumptions are satisfied? Explain

4. How many more students should be included in the sample of 90% sure that a point estimate p will be within a distance of 0.05 from p.

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

1)

Sample size = n = 78

x = 59

Sample proportion is

2)

We have to construct 90% confidence interval for the population proportion.

Formula is

Here E is a margin of error.

Zc = 1.64

So confidence interval is ( 0.7564 - 0.0797 , 0.7564 + 0.0797) => ( 0.6767 , 0.8361)

3)

Assumptions:

  • The sample is a simple random sample.
  • The normal distribution is used to approximate the distribution of sample proportions.

Yes, the assumptions are satisfied.

4)

Margin of error = E = 0.05

Confidence level = C = 0.90

Zc = 1.64   ( Using z table)

p = 0.7564

We have to find the sample size (n)

Add a comment
Know the answer?
Add Answer to:
A random sample of 78 students were interviewed and 59 said they would vote for a...
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
  • In a random sample of 500 U.S. citizens, 268 said that they plan to vote in...

    In a random sample of 500 U.S. citizens, 268 said that they plan to vote in the upcoming election. a. Construct a 90% confidence interval for the proportion of all U.S. citizens who would say they plan to vote in the upcoming election. b. Explain why the interval does or does not indicate that a majority of U.S. citizens would say they plan to vote in the upcoming election. c. Explain why the interval does or does not indicate the...

  • 3. Suppose a researcher believes that college faculty vote at higher rate than college students. She...

    3. Suppose a researcher believes that college faculty vote at higher rate than college students. She collects data from 200 college faculty and 200 college students using simple random sampling. If 167 of the faculty and 138 of the students voted in the 2008 Presidential election, (A) Find a 90% confidence interval for the difference between the proportion of faculty and the proportion of students who vote. (b) Is there enough evidence at the 3% level of significance to support...

  • A random sample of students found that 36 out of 190 were right handed. a. Let...

    A random sample of students found that 36 out of 190 were right handed. a. Let p represent the proportion of all students that are right handed. Find a point estimate for p. b. Find a 95% confidence interval for the population average p.

  • 26. In a random sample of 95 college students, 40 wished they would have chosen a...

    26. In a random sample of 95 college students, 40 wished they would have chosen a different major. Use the following steps to construct a 95% confidence interval for the true proportion of all students who wished they would have chosen a different major. a. Find the number of sample values, n b. Find the sample proportion, B c. Find the critical z-score, 2/2 d. When calculated correctly, E = 0.0993. Construct a confidence interval for the population proportion, p....

  • What proportion of JMU students like online courses? To estimate this, a random sample of 200...

    What proportion of JMU students like online courses? To estimate this, a random sample of 200 students was taken and found that 50 of them like online courses. (1) Are the assumptions for constructing a confidence interval for the population proportion satisfied? Explain.(2) Construct a 99% confidence interval for the population proportion of JMU students who like online courses.(3) What is the margin of error in (2) ?______________(4) Interpret the 99% confidence interval.

  • 2. Suppose that a random sample of 41 state college students is asked to measure the...

    2. Suppose that a random sample of 41 state college students is asked to measure the length of their right foot in centimeters. A 90% confidence interval for the mean foot length for students at this university turns out to be (21.709, 25.091). If we now calculated a 95% confidence interval, would the new confidence interval be wider than or narrower than or the same as the original? b. Suppose two researchers want to estimate the proportion of American college...

  • A researcher wants to use a confidence interval to estimate the proportion of college students in...

    A researcher wants to use a confidence interval to estimate the proportion of college students in his state who plan to vote in the 2020 presidential election. He plans to randomly sample 120 college students, and plans to construct a 95% or 99% confidence interval. Which of these confidence intervals will be wider, and why? Group of answer choices _99%. As the level of confidence increases, the width of the confidence interval increases. _95%. As the level of confidence decreases,...

  • A sample of college students was asked whether they would return the money if they found...

    A sample of college students was asked whether they would return the money if they found a wallet on the street. Of the 93 women, 84 said "yes," and of the 75 men, 47 said "yes." Assume that these students represent all college students. (a) Find separate approximate 95% confidence intervals for the proportions of college women and college men who would say "yes" to this question. (Round the answers to three decimal places.) for women for men (b) Find...

  • please answer all parts Q1 A school found that 78 out of a random sample of...

    please answer all parts Q1 A school found that 78 out of a random sample of 100 students had read a book in a month. (a) Find a 96% confidence interval for the proportion in the corresponding population. (b) The school has decided that a more precise estimate is needed and wishes to reduce the width of the confidence interval by a quarter. How many more students should be investigated if the school wants to find the 96% confidence interval...

  • A random sample of students were asked whether they were satisfied with their school's lunch menu....

    A random sample of students were asked whether they were satisfied with their school's lunch menu. The resulting confidence interval for the proportion of students who are satisfied is (0.33,0.51). What is the margin of error?

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