A poll conducted randomly in 2014 found that 52% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

A study of 40 English professors showed that they spend, on average, 12.6 minutes correcting a student's term paper. The standard deviation was 2.5 minutes. Construct the 99% confidence interval of the mean for all term papers.

An apartment hunter is shopping for either a one-bedroom or two-bedroom apartment in varying apartment buildings. She wants to estimate the additional amount she would have to pay, on average, for a two-bedroom apartment versus a one-bedroom apartment. The table shows the prices, in dollars, for one- and two-bedroom apartments from the nine buildings she visited.   A 95% confidence interval for the true mean difference between two-bedroom and one-bedroom apartments for apartment buildings like the ones in the sample was (255.93, 508.51). A friend of the apartment hunter tells her to expect to pay about $500 more per month for a two-bedroom apartment than for a one-bedroom apartment. Does the confidence interval provide evidence for the friend’s claim? A. No, the confidence interval does not contain 0, so the friend’s claim is not plausible. B. No, the confidence interval contains many more numbers below 500 than above 500, so it is unlikely that the friend’s assertion is correct. C.…

A research company is interested in determining the true proportion of Americans that work remotely. In a random sample of 200 individuals, 4.5% of them work remotely. Find a 90% confidence interval for the true proportion of Americans that work remotely.

Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300 said they were current smokers. d. Interpret the 95% confidence interval in this context.

Several years ago, 44% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,075 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,075 surveyed, 479 indicated that they were satisfied. Construct a 95% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed. What are the null and alternative hypotheses? Ho: p = 0.44 versus H,: p 0.44 (Round to two decimal places as needed.) Use technology to find the 95% confidence interval. The lower is The upper bound is (Round to two decimal places as needed.)

Suppose you calculate a 99% confidence interval of 14.71 to 21.64. Determine the margin of error.

A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 121 randomly selected Southerners were surveyed and found to watch a mean of 4.6 hours of sports per week. In the North, 176 randomly selected people were surveyed and found to watch a mean of 3.2 hours of sports per week. Find a 90% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.6 hours per week and the North has a population standard deviation of 1.4 hours per week. Let Population 1 be people who live in the South and Population 2 be people who live in the North. Round the endpoints of the interval to one decimal place, if necessary. Lower endpoint: Upper endpoint:

A learn-to-type software program claims that it can improve your typing skills. To test the claim and possibly help yourself out, you and six of your friends decide to try the program and see what happens. Use the table below to construct an 80% confidence interval for the true mean change in the typing speeds for people who have completed the typing program. Let Population 1 be the typing speed before taking the program and Population 2 be the typing speed after taking the program. Round the endpoints of the interval to one decimal place, if necessary. Typing Speeds (in Words per Minute) Before After 32 33 51 42 55 40 35 32 54 33 45 30 50 52

A survey of 32,782 adults in 2004 indicated that 24.6% of adults were current smokers. A similar study conducted in 1995 of 40,000 adults indicated that 29.7% were current smokers. a. Find and interpret a point estimate of the difference between the proportion of current smokers in 1995, and the proportion of current smokers in 2004. b. A 95% confidence interval for the true difference is (-0.058,-0.045). Interpret. c. What assumptions must you make for the confidence interval i part b to be valid? (…) a. Find a point estimate of the difference between the proportion of current smokers in 1995 and the proportion of current smokers in 2004. (Round to three decimal places as needed.)