12. The confidence interval for the true mean weight of the residents of the town is 152 ± 9 pounds.
13. The confidence interval for the true mean SAT score of the graduating high school seniors is 454 ± 27.
14. The margin of error on the survey is (83% - 75%) / 2 = 4%.
15. The margin of error on the survey is (173 - 159) / 2 = 7 pounds.
Determine Confidence Interval or Margin of Error
12. A study was commissioned to find the mean weight of the residents in certain town. The study found the mean weight to be 152 pounds with a margin of error of 9 pounds. Write a confidence interval for the true mean weight of the residents of the town.
13. A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study found that the mean SAT score was 454 with a margin of error of 27. Write a confidence interval for the true mean SAT score of the graduating high school seniors.
14. A survey was given to a random sample of the residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. The survey reported a confidence interval that between 75% and 83% of the residents supports the plan. What is the margin of error on the survey?
15. A study was commissioned to find the mean weight of the residents in certain town. The study found a confidence interval for the mean weight to be between 159 pounds and 173 pounds. What is the margin of error on the survey?
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