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- Virginia wants to estimate the percentage of students who live more than three miles from the school. She wants to create a 95% confidence interval which has a margin of error of at most 5%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Use the table of values above.Bert computes a 95% confidence interval for p and obtain the interval [0.600,0.700]. Bert boss says, Give me a 95% confidence interval for p. Calculate the answer for Bert.A poll of 2612 U.S. adults found that 15% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: _______% to ______% Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: ________% to ________% Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: ________% to ______% The more error we allow, the less precise our estimate. Therefore, as the confidence…
- A poll of 2368 U.S. adults found that 40% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our…Construct a 95% confidence interval for p1 - p2 for a survey that finds 30% of 240 males and 41% of 200 females are opposed to the death penalty. Group of answer choices a.(-0.200, -0.021) b.(-1.532, 1.342) c.(-1.324, 1.512) d.(-0.561, 0.651)A new fertilizer was applied to the soil of 239 bean plants. 11% showed increased growth. Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval:
- A poll of 2338 U.S. adults found that 71% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to %Supermercados VIVA's marketing manager wants to estimate the proportion of customers who would not recommend the brand. What should be the minimum sample size that you will need to perform this estimate, if you want results with a confidence level of 96.6% and a margin of error of 4.4%? 1. The formula to be used corresponds to the sample size to estimate a:2.The table value associated with the confidence level is:3.The number of clients selected to be part of the sample is:A new fertilizer was applied to the soil of 246 bean plants. 25% showed increased growth. Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places. Margin of Error (as a percentage):Confidence Interval: % to %
- Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n= 1004 and x = 575 who said "yes." Use a 90% confidence level. E Click the icon to view a table of z scores. .... a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.)98% (lype an integer or a decimal.) d. Il the claim in part (c) is tested using this sample data, we get this confidence interval: -0.000620Answer question 12 and show all the steps.SEE MORE QUESTIONS