A recent study claimed that 66% of the population of all college students prefer printed textbooks to electronic textbooks. You want to test this daim by surveying a random sample of 68 college students. Follow the steps below to construct a 95% confidence interval for the population proportion of all college students who prefer printed textbooks to electron textbooks. Then state whether the confidence interval you construct contradicts the study's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results from the random sample. Number Proportion Prefers printed textbooks to electronic Take Sample 51 0.75 textbooks Does not prefer printed textbooks to electronic textbooks 17 0.25 Enter the values of the sample size, the point estimate of the population proportion, and the critical value you need for your 95% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". Sample size: Standard error: Critical values Point estimate: F0.00s 2.576 Margin of error: F0.010 -2.326 Critical value: F0.025-1.960 95% confidence interval: -1.645

A survey of 240 teachers found that two-thirds of those surveyed do not live in Sydney. Can you conclude statistically that the population proportion who do not live in Sydney is less than ​0.75? Use a​ 95% confidence interval. What is the confidence interval and is 0.75 within the interval?

A beverage company would like to determine if their drink is preferred to a rival company’s drink. They perform an experiment with 150 randomly selected individuals and have them taste test the two different drinks in a controlled environment so that each person's preference is independent of others. A 90% confidence interval for the proportion preferring their drink is (0.453, 0.587). Which of the following is the correct interpretation of this confidence interval? 90% of all consumers prefer this beverage company’s drink to their rival’s between 45.3% and 58.7% of the time. We’re 90% confident that between 45.3% and 58.7% of all consumers would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that there’s a 45.3% to 58.7% chance that a randomly selected person would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that the proportion of people in the sample who prefer this beverage company’s drink over their…

A study is conducted by a physical therapist to determine whether there is a difference in the proportion of men and women in the U.S. that participate in regular sustained physical activity. It is found that 22% of menout of 550 and 19.5% of women out of 450 participated in regular sustained physical activity. Construct a 90% confidence interval for the difference in proportions of men and women that participate in regular sustained and state your conclusion.

A recent poll indicated that 42% of men supported raises taxes to fund additional park services.   The poll also found that 64% of women supported such a tax.  The sample consisted of 29 men and 25 women.   Construct a 90% confidence interval around the difference in proportions.

A random sample of 1700 car owners in a particular city found 816 car owners who received a speeding ticket this year. Using the Wilson "plus four" technique, find a 95% confidence interval for the true percent of car owners in this city who received a speeding ticket this year. Express your results to the nearest hundredth of a percent.

A marketing research company is estimating which of two soft drinks college students prefer. A random sample of n college students produced the following 95% confidence interval for the proportion of the college students who prefer to drink A:(0.342,0.482). Identify the point estimate for estimating the true proportion of college students who prefer that drink.