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.

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1. (0.719, 0.841) 2. (0.729, 0.831) 3. (0.712, 0.847) 4. (0.737, 0.823)

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

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1. True 2. False

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We have calculated a 95% confidence interval for p and would prefer to have a smaller margin of error

z

*

n

without losing any confidence. In order to do this, we can I. change the z ∗

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value to a smaller number.

II. take a larger sample.

III. take a smaller sample.

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1. I only 2. II only 3. III only 4. I and II

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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 to be?

z

*

n

fourth as large. How large does our new sample need 1. 25 2. 50 3. 200 4. 400 5. 1600

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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

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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)

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1. 32 2. 1418 3. 999 4. 38

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At Dartmouth College students can buy an 18 meals/week food plan or a 14 meals/week food plan. A campus organization

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calculated a 95% confidence interval for p, the proportion of students with a 14 meals/week plan, as (0.58, 0.66). Choose the

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1. We are 95% confident that the true proportion p of students with a 14 meals/week plan is between 0.58 and 0.66.

2. 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.

3. 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.

4. 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.

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