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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.

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Presentation on theme: "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."— Presentation transcript:

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

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

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 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

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? 10 25 50 200 400 1600

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) 10 32 1418 999 38

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. 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.


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