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(0.719, 0.841) (0.729, 0.831) (0.712, 0.847) (0.737, 0.823)

A survey found 195 of 250 randomly selected Internet users have high-speed Internet access at home. Construct a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. (0.719, 0.841) (0.729, 0.831) (0.712, 0.847) (0.737, 0.823). 10.

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(0.719, 0.841) (0.729, 0.831) (0.712, 0.847) (0.737, 0.823)

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  1. A survey found 195 of 250 randomly selected Internet users have high-speed Internet access at home. Construct a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. • (0.719, 0.841) • (0.729, 0.831) • (0.712, 0.847) • (0.737, 0.823) 10

  2. The higher the level of confidence we want, the narrower our confidence interval becomes. • True • False 10

  3. We have calculated a 95% confidence interval for p and would prefer to have a smaller margin of error without losing any confidence. In order to do this, we canI. change the z∗ value to a smaller number.II. take a larger sample.III. take a smaller sample. • I only • II only • III only • I and II 10

  4. We have calculated a confidence interval based on a sample of size n = 100. Now we want to get a better estimate with a margin of error that is only one-fourth as large. How large does our new sample need to be? • 25 • 50 • 200 • 400 • 1600 10

  5. A news poll which estimated that 82% of all voters believe global warming exists had a margin of error of +/- 3%. Suppose an environmental group planning a follow-up survey on this issue wants to determine a 95% confidence interval with a margin of error of no more than 2%. How large a sample do they need? (For estimate of p use 0.82) • 32 • 1418 • 999 • 38 10

  6. At Dartmouth College students can buy an 18 meals/week food plan or a 14 meals/week food plan. A campus organization calculated a 95% confidence interval for p, the proportion of students with a 14 meals/week plan, as (0.58, 0.66). Choose the correct interpretation of this confidence interval. • We are 95% confident that the true proportion p of students with a 14 meals/week plan is between 0.58 and 0.66. • In 95% of all random samples of Dartmouth students, the sample proportion p of students with a 14 meals/week plan is between 0.58 and 0.66. • We are 95% confident that the interval (0.58, 0.66) has captured the true proportion p of students with a 14 meals/week plan. • In 95% of all random samples of Dartmouth students, the sample proportion p of students with a 14 meals/week plan will be 0.62. 10

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