question

TSE=? You have determined that a sample size of 250 will allow reasonable precision and confidence for your estimates of important population parameters. You will be conducting a telephone interview of university undergraduate students. After checking with university registration officials, you determine the campus directory to be 85% accurate. Further, you expect about 2% of the people you contact to be ineligible (e.g., graduate students), about 20% to decline your participation request, and about 30% not to be reachable even after trying at several different times on different days of the week. How many sampling elements should you include in the project?

answer

250/(1-BCI)(1-I)(1-R)(1-NC) =250/(1-.15)(1-.02)(1-.2)(1-.3) =536

question

determining sample size(公式)

answer

n=(z^2/H^2)(est σ^2)

question

Determine Sample Size: You have been asked to determine the average amount that fishermen spend per year on food and lodging while on fishing trips in a certain state. Your estimate is to be within +/- $25 of the population mean, the confidence level is to be 95%, and the estimated standard deviation for the amount spent is $125 based on prior research.

answer

H=25 z=1.96 σ^2=125^2

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Determining Sample Size When Estimating Proportions(公式)

answer

n={z^2/H^2)π(1-π)

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Determining Sample Size When Estimating Proportions You have been asked to determine the proportion of all out-of-state fishermen who took at least one overnight fishing trip in the past year. Your estimate is to be within +/- 2% of the population mean, the confidence level is to be 95%, and the best guess is that 25% of out-of-state respondents have taken at least one overnight fishing trip.

answer

H=2% z=1.96 π=25%

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question

Response rates You've conducted an email survey, sending out 500 surveys; the 500 sample elements have been classified as follows: Usable questionnaires returned: 250 nSurveysSent: 500 Estimated eligibility percentage: .85

answer

250/(.85*500)=0.588

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question

What is the sampling error for means? Researchers learned that the average number of visits to Avery Fitness over the last 30 days was 10 visits with a standard deviation of 7.3 from 198 responding members We would like to establish a 95% confidence level (z = 1.96)

answer

1.96 (7.3/√198) = 1.0

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question

What is the confidence interval? Researchers learned that the average number of visits to Avery Fitness over the last 30 days was 10 visits with a standard deviation of 7.3 from 198 responding members We would like to establish a 95% confidence level (z = 1.96)

answer

10.0 - 1.0 ≤ x ≤ 10.0 + 1.0

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