Question

In: Statistics and Probability

A simple random sample of 300 colleges and universities was selected and for each it was...

A simple random sample of 300 colleges and universities was selected and for each it was determined whether they planned to move summer 2020 classes online or not. Of the 300 colleges and universities in the sample, 90 responded that they will move summer 2020 classes online , and 210 responded that they will not move summer 2020 classes online. . If appropriate, use this information to calculate and interpret a 99% confidence interval for the proportion of all colleges and universities that will move their summer 2020 classes online . Please type your solution in the box below.

Solutions

Expert Solution

Solution :

Given that,

n = 300

x = 90

Point estimate = sample proportion = = x / n = 90 / 300 = 0.30

1 - = 1 - 0.30 = 0.70

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 (((0.30 * 0.70) /300 )

= 0.068

A 99% confidence interval for population proportion p is ,

± E

= 0.30  ± 0.068

= ( 0.232, 0.368 )

We are 99% confident that the true proportion of all colleges and universities that will move their summer 2020 classes online between 0.232 and 0.368.


Related Solutions

A simple random sample of 80 colleges and universities is selected, and 16 indicated will move...
A simple random sample of 80 colleges and universities is selected, and 16 indicated will move their summer 2020 classes online while the other 64 indicated they did not plan to move their summer 2020 classes online . If appropriate, use this information to test the hypotheses stated in question 10 at the a = .10 level of significance. Please type your answer in the box below. Hypothesis: H0:   π = 0.14 versus HA: π > 0.14
A simple random sample of 35 colleges and universities in the United States has a mean...
A simple random sample of 35 colleges and universities in the United States has a mean tuition of 18700 with a standard deviation of 10800. Construct a 99% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.  
A simple random sample of 40 colleges and universities in the United States has a mean...
A simple random sample of 40 colleges and universities in the United States has a mean tuition of 17800 with a standard deviation of 11000. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.
A simple random sample of 40 colleges and universities in the United States has a mean...
A simple random sample of 40 colleges and universities in the United States has a mean tuition of 18,800 with a standard deviation of 10,100. Construct a 98% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.
A simple random sample of 101 cities and counties in the United States was selected and...
A simple random sample of 101 cities and counties in the United States was selected and the number of positive COVID-19 cases per 1000 citizens was recorded for each. The mean number of positive cases for this sample of 101 counties and cities was 253, with a standard deviation of 11.4. Due to a couple of cities with extremely high numbers of cases the distribution is skewed heavily to the right. If appropriate, use this information to calculate and interpret...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 20 days and Parsnip was fed pellets on the other 17 days. The goal is to calculate a 95% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
A simple random sample of 81 is selected from a population with a standard deviation of...
A simple random sample of 81 is selected from a population with a standard deviation of 17. The degree of confidence is 90%. What is the margin of error for the mean?
A sample of College Graduates from Rhode Island Colleges and Universities was taken where men and...
A sample of College Graduates from Rhode Island Colleges and Universities was taken where men and women were asked when they figured out what their college major was going to be. Use the data being represented in the table to answer the following questions. Before High School High School Freshman Year Sophomore Year Totals Male 4 22 56 37 119 Female 12 29 42 31 114 Totals 16 51 98 68 233 a. (4 pts.) What is the probability you...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT