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.

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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. 10 (0.719, 0.841) (0.729, 0.831) (0.712, 0.847) (0.737, 0.823)

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

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 can I. change the z∗ value to a smaller number. II. take a larger sample. III. take a smaller sample. 10 I only II only III only I and II

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? 10 25 50 200 400 1600

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) 10 32 1418 999 38

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. 10 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.