Construct a 95% confidence interval estimate for the percentage of students who prefer online learning.
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- Of 300 randomly selected medical students, 30 said that they planned to work in a rural community. Find 90% confidence interval for the true proportion of all medical students who plan to work in a rural community. The answers below represent approximate values.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 college students who prefer drink A: (.262, .622). Identify the point estimate for estimating the true proportion of college students who prefer that drink.A) .622 B) .442 C) .262 D) .18Refer to the accompanying data set of 20 randomly selected presidents. Treat the data as a sample and find the proportion of presidents who were taller than their opponents. Use that result to construct a 95% confidence interval estimate of the population percentage. Based on the result, does it appear that greater height is an advantage for presidential candidates? Why or why not? Click the icon to view the table of heights. Construct a 95% confidence interval estimate of the percentage of presidents who were taller than their opponents. (Round to one decimal place as needed.) McKinley 170 178Lincoln 193 188Reagan 185 177Taft 182 178Harrison 173 168Van Buren 168 180Harding 183 178Eisenhower 179 178Taylor 173 174Garfield 183 187J. Kennedy 183 182Harrison 168 180J. Q. Adams 171 191Clinton 188 188Nixon 182 180Johnson 192 180Hoover 182 180T. Roosevelt 178 175Buchanan 183…
- In each of Exercises, use the one-proportion plus-four z-interval procedure to find the required confidence interval. Interpret your results. Working with Millions. A poll by Gallup asked, “If you won 10 million dollars in the lottery, would you continue to work or stop working?” Of the 1039 American adults surveyed, 707 said that they would continue working. Obtain a 95% confidence interval for the proportion of all American adults who would continue working if they won 10 million dollars in the lottery.In a simple random sample of 300 college students, 35.0 percent stated that they intended to pursue an advanced degree. [Show Work] A. How many of the 300 college students sampled intend to pursue an advanced degree? B. To the 95 percent confidence level, find the confidence interval estimate, (to nearest tenth of a percent) of ALL college students who intend to pursue an advanced degree, Can we safely assume that less than 40.0% of all college students intend to pursue an advanced degree? C. If we wanted to reduce the margin of error to 4.0 percent, how big would our sample SIZE have to be (again, to the 95 percent confidence level)?A collage professor would like to estimate the proportion of students who pull an *all-nighter," meaning they study all night for an upcoming exam. She selects a random sample of 100 students from her large college and finds that the 99% confidence interval for the true proportion of students who have pulled an all-nighter to be 0.48 to 0.62. If the professor had randomly selected 50 students rather than 100 students, what effect would this have had on the width of the interval? O it would have doubled. O It would have been cut in half. O it would have remained the same. O it would have been larger, but it would not have doubled.
- The Anthropology Department at UC Berkeley is measuring student heights. 100 male students are selected at random who show an average height of 180 cm, with an SD of 5 cm. Calculate a 95%-confidence interval for the average height of all UC Berkeley male students, rounded to the nearest decimal. a. (178cm, 182cm) b. (179cm, 181cm) c. (172 cm, 188 cm) d. (175cm, 185cm)From a survey of 400 new Wawasan Open University students chosen at random, it was discoveredthat 240 students have been to the library every day. Using this data, construct(a) 95% confidence interval (b) 99% confidence interval for the proportion of students who use the library every dayA consumer group reports that a 90% confidence interval for the mean fat content of chicken nuggets from a fast-food restaurant is (14, 18). What is the mean fat content for the sample of 30 chicken nuggets or the point estimate that help make this confidence interval? A. 32 B. 16 C. 2
- Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 99% confidence interval for the true proportion of all medical students who plan to work in a rural community. a) Confidence Interval b) ConclusionThe U.S. Department of the Interior is checking cattle on the Windgate open range in Montana. Arandom sample of 900 shows that 54 are undernourished. a) Let p represent the proportion of undernourished cattle on Windgate range. Find a point estimatefor p.b) Find a 99% confidence interval for p.If n=490 and ˆpp^ (p-hat) =0.54, find the margin of error at a 90% confidence levelGive your answer to three decimals